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Exam Tennis Serve 1. A flexed elbow at the start of the execution phase: A flexed elbow at the start of the execution phase in a tennis serve positions the arm closer to the body, reducing the moment of inertia. τ = Iα, a smaller I allows for greater angular acceleration (α), enabling a faster...

Exam Tennis Serve 1. A flexed elbow at the start of the execution phase: A flexed elbow at the start of the execution phase in a tennis serve positions the arm closer to the body, reducing the moment of inertia. τ = Iα, a smaller I allows for greater angular acceleration (α), enabling a faster arm and racquet speed to increase serve velocity. 2. Extension of the elbow during late execution as the racquet head hits the ball: Extending the elbow during the late phase of the tennis serve increases the linear velocity of the racquet head to the angular velocity of the arm (v = ω ⋅ r), where ω is angular velocity and r is the distance from the shoulder to the racquet head. This extended lever maximises racquet speed at impact, transferring more momentum to the ball and improving serve power and effectiveness. 3. Sequential movement of the trunk and upper limb segments: The sequential movement of the trunk and upper limb segments in a tennis serve utilises the kinetic chain, allowing energy to transfer progressively from the larger, slower trunk muscles to the smaller, faster muscles in the arm. This principles maximises velocity at the hand by the time of ball contact. v = ω ⋅ r, where increasing angular velocity (ω) of each segment amplifies the final hand velocity (v), resulting in a more powerful serve. Baseball Batting 1 1. Side on stance in set up: Using a side-on stance in the set-up increases the batter's rotational range of motion (θ) around the vertical axis. ω = θt , this greater angular displacement over the same swing time results in a higher angular velocity of the bat. The increased angular velocity enhances bat speed at impact, improving momentum transfer to the ball and the effectiveness of the hit. 2. Feet shoulder width apart and weight distributed more on the back foot in set up: Positioning the feet shoulder-width apart with weight on the back foot enhances stability and sets up for effective weight transfer during the swing. This weight shift forward increases linear and angular momentum (p = mv and L = Iω) leading to a higher bat speed at impact. The greater momentum results in a more powerful force transfer to the baseball. 3. Leaning the trunk backwards away from the bat during the hitting motion: Leaning the trunk backward away from the bat during the hitting motion increases the torso's rotational range of motion, resulting in a greater angular displacement (θ). ω = θt , this larger angular displacement over the same time period (t) yields a higher angular velocity (ω). The increased angular velocity enhances bat speed at impact. Baseball Batting 2 1. Stepping into the pitch during the hitting action: Stepping into the pitch increases the batter's linear momentum toward the ball (p = mv). This added momentum is transferred to the bat during the swing, enhancing the force applied to the ball upon impact. According to the conservation of momentum, this results in a higher exit velocity of the ball. 2. Keeping the elbows flexed and the bat close into the trunk during the early part of the swing (execution): Keeping the elbows flexed and the bat close to the trunk reduces the bat's moment of inertia (I) by bringing its mass closer to the body's rotational axis. τ = Iα (where τ is torque and α is angular acceleration, a smaller I allows for a greater α given the same torque. This increased angular acceleration enables the batter to achieve a higher bat speed at impact. 3. Extending the elbows during the execution of the swing: Extending the elbows during the swing increases the radius (r) of rotation of the bat around the body. v = rω, where v is linear velocity and ω is angular velocity, a larger r results in a higher bat speed at the point of contact. This increased bat speed enhances the momentum transferred to the baseball (p = mv). Sprinting 1. Using biomechanical principles, explain why sprinters are better able to accelerate when using blocks compared to when using a standing start? Sprinters can generate greater horizontal force and acceleration form blocks due to an increased angle of push-off and optimised application of ground reaction force (F = ma), aligning force more horizontally. Blocks provide resistance for a powerful backward push, maximising forward acceleration by increasing the initial ground reaction force vector in the direction of desired motion. 2. Elite sprinters tend to have more knee flexion during the swing phase than novice sprinters. Explain biomechanically how this improves sprint performance? Elite sprinters have more knee flexion during the swing phase to reduce the moment of inertia (I = mr 2 ) of the swinging leg by decreasing the radius (r) from the hip joint to the lower leg. This allows for faster hip rotation and quicker leg recovery, which increases stride frequency and improves sprints performance. 3. If a heavier sprinter and lighter sprinter exert the same force on the blocks explain which sprinter will have the greater acceleration: The lighter sprinter will have greater acceleration because, according to Newton's second law, F = ma or rearranged as a = m F , acceleration (a) is inversely proportional to mass (m) when force (F ) is constant. With less mass, the lighter sprinter achieves a higher acceleration for the same applied force. Horizontal Jump 1. Arm swing during the execution phase of the jump: Arm swing during the execution phase of a horizontal jump increases the impulse applied to the ground, enhancing take-off velocity. J = F ⋅ t = Δp, where J is impulse and Δp is the change in momentum, the additional force and time from swinging the arms results in a greater Δp. This increased momentum improves jump effectiveness by maximising horizontal distance and contributes to safety through better control during take-off and landing. 2. Hip and knee flexion during the preparation phase of the jump: Hip and knee flexion during the preparation phase increases the range of motion and distance over which force is applied. Work-energy principle (W ork = F orce ⋅ Distance), this greater distance allows more work to be done, resulting in increased kinetic energy (KE = 12 mv 2 ) at takeoff. This enhances the effectiveness of the horizontal jump by increasing jump distance and improves safety by distributing forces more evenly across joints. 3. Forward trunk lean prior to take-off: Leaning the trunk forward prior to take-off shifts the body's centre of mass ahead of the support foot, increasing the horizontal component of the ground reaction force (F horizontal ). Impulse = F ⋅ t = m ⋅ Δv, this greater F horizontal over the take-off time (t) enhances horizontal momentum (mv), resulting in a longer jump distance. The forward trunk alignment helps control rotational inertia, improving safety by ensuring a more stable landing. 4. Hip and knee flexion on landing from the jump: Hip and knee flexion upon landing increases the time (t) over which the body decelerates. F ⋅ t = m ⋅ Δv, increasing t reduces the impact force (F ) experienced. This reduction in force minimises stress on joints and muscles, enhancing safety and effectiveness during the horizontal jump. Ice Skater  Explain biomechanically why the skater flexes her elbows and adducts her hips at the start of the spin and then does the opposite actions to finish the spin? By flexing her elbows and adducting her hips at the start of the spin, the skater reduces her moment of inertia (I) by bringing mass closer to the axis of rotation. Conservation of angular momentum, L = Iω, a decrease in I results in an increase in angular velocity, allowing her to spin faster. To finish the spin, she does the opposite, extending her limbs increases I, which decreases ω and slows down the spin. Throwing Technique Comparison Describe one difference in the wind-up phase (preparation phase) between the adult and the child: The adult demonstrates greater trunk rotation and shoulder external rotation, allowing for increased storage of elastic energy and greater torque generation. The child technique involves less trunk rotation and external rotation due to limited strength and motor control. Explain biomechanically why the difference you have noted in (a) is likely to result in the adult being able to generate greater ball velocity at release compared to the child? An adult's greater trunk rotation and shoulder external rotation increase torque generation (τ = Iα) and allow for more elastic energy storage during the wind-up. This results in higher angular acceleration (α) and angular velocity, leading to greater ball speed at release (v = rω). Describe one difference in the execution phase of the throw between the adult and the child: The adult uses a sequential kinetic chain during the execution phase, starting with the legs and hips to generate power, followed by the trunk and arm, optimising energy transfer. In contrast, the child uses a single-segment movement pattern, relying primarily on the arm, which reduces throwing velocity and efficiency due to limited use of the lower body and trunk. Explain biomechanically why the difference you have noted in (c) is likely to result in the adult being able to generate greater ball velocity at release compared to the child? An adult's use of a sequential kinetic chain allows for efficient energy transfer from the legs and hips through the trunk to the arm, maximising ball velocity at release. According to v = ωr, the combined angular velocities of each segment enhance the final speed of the ball. In contrast, a child relying primarily on arm movement lacks this cumulative energy transfer, resulting in a lower ball velocity at release. Bicep Curl Is the biceps muscle at a mechanical advantage or disadvantage during the bicep curl? Explain why: The biceps muscle is at a mechanical disadvantage during a biceps curl because it functions as a third-class lever, where the effort (muscle insertion) lies between the fulcrum (elbow joint) and the load (weight in hand). The mechanical advantage is calculated by MA = LoadArm ; since the effort arm is shorter than the load arm, MA is EffortArm less than 1. This means the biceps must produce a force greater than the load to lift it, resulting in a mechanical disadvantage. Is the external resistance torque produced by the dumbbells greater at the start of the curl or at 90 degrees of elbow flexion? Explain why: At 90° of elbow flexion during a biceps curl, the moment arm (d ⊥ ) of the weight relative to the elbow joint is maximised. τ = F ⋅ d ⊥ , where F is the force due to the weight and d ⊥ is the perpendicular distance from the joint to the line of action of the force, a larger d ⊥ at 90° results in greater external resistance torque. Therefore, the torque is greater at 90° compared to the start of the curl, where the moment arm and thus the torque are smaller. Gymnast Why is it more difficult to balance while walking on the beam than it is walking on a footpath? Balancing on a beam is more difficult than on a footpath because the narrow base of support reduces stability, making it easier for the centre of mass to shift outside the support area, which increases the likelihood of tipping over. Stability is maximised when the centre of mass remains within the base of support, and with a smaller base, even slight shifts result in greater instability. What does the gymnast do to regain balance after the jump? By sticking out her arm and leg, the gymnast increases her moment of inertia ( I = ∑ mr 2 ) around her centre of mass. According to the conservation of angular momentum (L = Iω), increasing I decreases angular velocity (ω), allowing her to slow down any unwanted rotation. This adjustment helps her counteract imbalance and regain stability upon landing. Why might a shorter gymnast have enhanced balance on the beam compared to a taller gymnast? A shorter gymnast has a lower centre of mass and a smaller moment of inertia (I = mk 2 ), where m is mass and k is the radius of gyration. According to α = τI (angular acceleration α equals torque τ divided by moment of inertia I), a smaller I allows for greater angular acceleration, enabling the gymnast to make quicker adjustments to maintain balance. Therefore, the shorter gymnast can more effectively control her body's movements on the beam, enhancing balance compared to a taller gymnast. Kicking (Punting) 1. Outstretched arm on the side opposite the kicking leg during the execution: An outstretched arm opposite the kicking leg increases the body's rotational inertia (I = m ⋅ r 2 ), counterbalancing the rotational forces generated by the kick. This stabilisation reduces unwanted torso rotation, allowing more of the force to be directed into the kicking leg, thereby improving the power and accuracy of the kick. 2. Backward lean of the trunk and extension of the knee of the kicking leg during execution: Leaning the trunk backward and extending the knee of the kicking leg increases the leg's range of motion and angular displacement (θ). ω = θt , this greater angular displacement over the same time period results in a higher angular velocity. This increased angular velocity leads to a higher foot speed at impact, enhancing force transfer to the ball and improving the effectiveness of the kick. 3. The kicking leg trails well behind the body during the early execution phase of the kick: Allowing the kicking leg to trail well behind the body increases the angular displacement (θ) through which the leg can accelerate. ω 2 = 2αθ, a larger θ results in a higher final angular velocity (ω) for a given angular acceleration (α). This increased angular velocity enhances foot speed at impact, improving momentum transfer to the ball and thus the effectiveness of the kick. 4. During early execution the hip of the kicking leg flexes and the knee remains flexed: Flexing the hip while keeping the knee flexed reduces the leg's moment of inertia (I) about the hip joint, bringing mass closer to the axis of rotation. τ = Iα, for a given torque (τ), a lower I results in a higher angular acceleration (α). This enables the kicking leg to accelerate more rapidly, increasing the foot speed at impact and enhancing the effectiveness of the kick. 5. The kicking knee is extended when contact is made with the ball: Extending the kicking knee at the moment of contact increases the radius (r) from the hip joint to the foot, enhancing foot speed according to v = rω, where v is linear velocity and ω is angular velocity. With a larger r and maintained ω, the foot's linear velocity v is higher at impact. This higher foot speed results in more effective momentum transfer to the ball. 6. The kickers body moves forward past the supporting foot in the direction of the kick: By moving the body forward past the supporting foot, the kicker increases linear momentum (p = mv) in the direction of the kick. This forward motion adds to the foot's velocity at impact, enhancing the transfer of momentum to the ball. According to the conservation of momentum, the increased body momentum results in a more powerful kick. Baseball Pitch 1. Large range of motion of the shoulder during the wind-up (preparation phase): A large range of motion of the shoulder during the wind-up increases the time and distance over which force is applied, maximising work (W = F × d) on the ball. This extended range allows the pitcher to generate greater angular velocity by the release point, increasing the balls velocity. 2. Large step toward the plate during the execution of the pitch: A large step toward the plate during a pitch increases the pitcher's stride length, which extends the distance over which force is applied, effectively increasing the work done (W = F d) on the baseball. This longer stride enables  greater forward momentum (momentum = mv), allowing for a higher release velocity, thus improving pitch speed. 3. Sequential motion of the segments during the execution of the pitch: The sequential motion of body segments in a baseball pitch enables each segment to reach peak velocity and transfer momentum to the next, maximising force application at the hand and ball. By starting with larger, slower segments (hips and torso) and moving to smaller, faster segments (shoulder and wrist), the pitch effectively increases velocity through a summation of forces, optimising v = r × ω for maximum ball speed. 4. Long follow through after ball release: A long follow-through after ball release reduces the deceleration forces on the arm, allowing a smother reduction in velocity (a = Δv Δt ) and minimising strain on the muscles. By extending the follow-through distance, more time is available for the arm to decelerate, reducing the peak forces experiences and allowing for greater control and consistency in achieving high velocities before release. Biomechanically, identify one positive and one negative of using a heavier bat when hitting in baseball? Using a heavier bat increases the potential to generate greater momentum (p = mv), allowing for more forceful impacts with the ball and potentially longer hits. However, a heavier bat also requires more force to accelerate (F = ma), which can reduce swing speed and control, especially during fast pitches, potentially hindering reaction time and precision. Split Squat Describe two biomechanical principles that contribute to the split squat on the left side (weights held level with shoulders) being less stable than the one on the right (weights held by the side): Holding weights at shoulder level during a split squat raises the centre of mass decreasing stability because a higher COM increases the torque (τ = F ⋅ d) generated by any lateral deviation, where d is the height of the COM above the pivot point. This position increases the body's moment of inertia (I = mr 2 ) with r being the distance from the axis of rotation, making it more difficult to control rotational movements. Therefore, both the elevated COM and increased moment of inertia contribute to reduced stability compared to holding weights Vertical Jump Describe how flexing the hips and knees before take-off can improve impulse production during the vertical jump: Flexing the hips and knees before take-off increases the duration over which force can be applied to the ground, enhancing impulse (Impulse = F × Δt). This extended time allows for greater momentum change, resulting in a higher jump by maximising the force exerted against the ground. Explain biomechanically why swinging the arms during the execution of the jump can improve vertical jump height: Swinging the arms during a jump generates additional upward momentum and increases the force applied over time, enhancing impulse (Impulse = F × Δt), which contributes to greater vertical takeoff velocity. This added momentum from the arms also increases kinetic energy (KE = 12 mv 2 ), allowing the jumper to reach a higher peak height. Explain how increased hip and knee flexion when landing from the jump can reduce ground reaction forces: Increased hip and knee flexion during landing extends the time over which the ground reaction force is applied, thus reducing the average force exerted on the body ( Impulse = F × Δt). By lengthening the deceleration time, the impulse required to bring the body to rest is spread out, lowering peak ground reaction forces and reducing the impact on joints. Hammer Throw The hammer thrower completes a number of rotations before she releases the hammer. Describe the biomechanical basis for why she rotates AND why she moves forward during the throw: The hammer thrower rotates to generate angular momentum, which maximises tangential velocity (v t = r ⋅ ω), where r is the radius (arm length plus cable) and ω is the angular velocity. Forward movement occurs due to ground reaction forces acting against the hammer's centripetal force, helping maintain balance and enabling a smooth, powerful release trajectory. Explain biomechanically why having the arms fully extended during the release of the hammer can improve throwing distance: Having the arms fully extended during release maximises the radius (r) in the tangential velocity equation. A larger radius increases tangential velocity, which directly enhances the hammer's release speed and therefore improves throwing distance due to greater kinetic energy (KE = 12 mv 2t ). Explain biomechanically why the hammer thrower leans backward during the throw: The hammer thrower leans backward to counterbalance the hammer's centripetal force (F c = m ⋅ ω 2 ⋅ r), where m is the hammer's mass, ω is angular velocity, and r is the radius of rotation. This backward lean lowers the centre of mass and aligns the body against the outward pull, increasing stability and control over the hammers path. What are the three main determinants of how far the hammer will travel horizontally? The three main determinants of hammer throw distance are release velocity, release angle, and release height. d = v 2 sin(2θ) g , where v is release velocity, θ is release angle, and g is gravitational acceleration. A higher release point also increases flight time. Disregarding increasing the height of release, describe two other advantages that tall athletes generally have for hammer throwing: Tall athletes generally have a longer radius from the centre of rotation to the hammer (r), allowing for a higher tangential velocity (v t = r ⋅ ω) at release, assuming equal angular velocity (ω). Additionally, they tend to have a larger moment of inertia (I = mr 2 ), helping stabilise their rotation and resist perturbations during the throw. Bench Press To be effective is the bench press a simultaneous or sequential movement? Why? The bench press is a simultaneous movement because the main muscle groups (chest, shoulders, triceps) contract together to generate maximum force, allowing for efficient force application directly upward. This coordination produces a constant force throughout the lift, maximising power output (P = F ⋅ v) as the bar is pressed away from the body. Dumbbell Lateral Raise 1. During the dumbbell lateral raise if the arm is held stationary in three positions (90° abduction, 45° abduction and vertically downwards), state which of these positions would be hardest to hold and why? Holding the dumbbell at 90° abduction is the hardest because torque ( T = F × d ⊥ ) is maximised when the moment arm (d ⊥ ) is longest. At 90°, the perpendicular distance from the shoulder joint to the weight is greatest, requiring more force to counterbalance the torque compared to positions at 45° or vertically downwards, where d ⊥ is shorter. 2. When the athlete holds the dumbbell with the shoulder in 90° abduction, how does this change his centre of mass (gravity) position compared to holding it by his side? When the athlete holds the dumbbell at 90° shoulder abduction, the centre of mass shifts further from the body's mid-line, increasing the moment arm (d ⊥ ) and creating a larger torque (T = F × d ⊥ ) around the shoulder. This requires greater muscular force to maintain the position, as the weight now exerts a stronger downward pull at this extended position. 3. Describe one way that he could alter his body position to improve balance while holding the dumbbell in this position. To improve balance during a dumbbell lateral raise, the individual could widen their stance, increasing his base of support. This adjustment helps aligns his centre of gravity over the base, reducing the risk of tipping due to the external torque created by the dumbbell's weight acting at a distance from his shoulder joint. Diving  1. Forward lean at the start of the dive: The forward lean at the start of the dive lowers the athlete's centre of mass and shifts it closer to the edge of the platform, reducing the torque needed to overcome static equilibrium. This setup allows the diver to generate greater forward angular momentum ( L = I ⋅ ω) with less initial effort, aiding in a faster and more effective dive by maximising initial horizontal velocity. 2. Tucking up the arms and legs during the middle of the dive: Tucking up the arms and legs during a dive reduces the diver's moment of inertia ( I = mr 2 ). According to the conservation of angular momentum (L = I ⋅ ω), decreasing I leads to an increase in rotational velocity ω. This faster spin enables the diver to complete more rotations before extending their body to slow down for a controlled entry into the water. 3. Extending out the arms and legs just prior to entry into the water: Extending the arms and legs before water entry increases the diver's moment of inertia (I = mr 2 ), slowing angular velocity due to the conservation of angular momentum (L = I ⋅ ω). This reduction in rotational speed helps the diver achieve a straighter, more controlled entry, minimising splash and aligning the body for streamlined water entry. Cricket Bowling 1. Running in prior to delivery of the ball: Running prior to delivery in fast bowling increases the bowler's linear momentum (p = mv), which can be transferred into rotational and translational energy during the bowling action, enhancing ball release velocity. This approach also allows for greater angular velocity of the arm during the delivery phase, maximising the final speed imparted to the ball. 2. Backward trunk lean during the delivery stride: A backward trunk lean during the delivery stride lengthens the path over which force can be applied, allowing the bowler to generate greater forward momentum. This increases the angular velocity of trunk rotation, which, due to the kinetic chain, transfers energy efficiently from the body to the arm and ball, maximising release velocity. 3. Keeping the delivery arm straight (elbow extended) at the time of ball release: Keeping the delivery arm straight at ball release maximises the radius (r) from the shoulder joint to the ball, increasing tangential velocity (v t = r ⋅ ω) if angular velocity (ω) is maintained. This extension also reduces energy loss from elbow flexion, enabling more efficient transfer of momentum to the ball for a faster release. Javelin Throw 1. Accelerating on the run up prior to the throw: Accelerating on the run-up increases the javelin thrower's linear momentum (p = m ⋅ v), which can be transferred into the throw at release, enhancing the javelin's launch speed. According to the principle of conservation of momentum, maximising the run-up velocity allows more kinetic energy to be imparted to the javelin, resulting in greater distance. Higher initial speed enables a more effective summation of forces from the legs through the torso and arms, optimising projectile velocity. 2. Rotating the pelvis and trunk away from the direction of the throw in the preparation phase: Rotating the pelvis and trunk away from the throw direction in the preparation phase increases the range of motion and stores elastic potential energy in the core and shoulder muscles, enhancing the force generated during the throw. This rotation allows for greater angular momentum (L = I ⋅ ω) to be transferred through the kinetic chain, leading to a higher release velocity of the javelin and improved throw distance. 3. Leaning backwards during the preparation phase: Leaning backward during the preparation phase increases the potential energy and optimises the stretch of the torso and shoulder muscles, enhancing the store of elastic energy. This position maximises the torque (τ = F ⋅ d) applied by the shoulder and core muscles when transitioning into the throw, allowing for a greater forward acceleration and release velocity of the javelin. 4. The throwing hand and javelin trailing behind the upper arm and trunk during the throwing action: The trailing of the throwing hand and javelin behind the upper arm and trunk during the throw maximises the stretch in the shoulder and trunk muscles, enhancing elastic energy storage and release (elastic recoil). This increase the angular velocity of the arm during the throw, contributing to a higher release velocity. As a result, this delay allows greater torque generation due to the increased range of motion (τ = I ⋅ α). 5. Large follow-through after the throw is complete: A large follow-through after a javelin throw reduces the deceleration forces acting on the arm, minimising injury risk and maximising the throw's distance. By extending the motion, the thrower allows momentum (p = m ⋅ v) to dissipate gradually, avoiding an abrupt stop that would exert high stress on joints. This technique aligns with the impulse-momentum relationship (impulse = Δp), where extending time reduces peak forces during deceleration. Power Clean 1. Shoulder width stance and shoulder width grip, with the grip of each hand the same distance from the plates: In a power clean, a shoulder- width stance and shoulder-width grip align the body's centre of mass with the barbell, optimising balance and stability. This setup ensures an efficient force transfer by allowing the bar to travel vertically close to the body, minimising the torque (τ = F ⋅ d) around the shoulder and hip joints. Keeping the grip equidistant ensures symmetric load distribution, reducing asymmetrical stress on the shoulders and spine. 2. Semi-squat position at set-up prior to execution: The semi-squat position in the power clean setup improves effectiveness by pre-loading the lower body muscles, allowing for optimal force generation through the stretch-shortening cycle. This position enables the athlete to apply greater ground reaction force, by increasing hip and knee flexion, which improves vertical displacement and power output (F = m ⋅ a). The semi-squat also positions the body's centre of mass directly over the base of support, enhancing the stability and control during the explosive lift phase. To be effective, is the power clean a simultaneous or sequential movement? Why? The power clean is a sequential movement because it requires a series of timed, coordinated actions to transfer force effectively from the ground to the barbell. The movement sequence progresses from the legs to the hips and finally to the upper body, allowing optimal force summation through each segment. This timing allows for maximal power generation (P = F ⋅ v). Biomechanically, why is lifting the bar too far out in front during execution of the lift not the most effective technique? Lifting the bar too far out in front during a lift creates a longer moment arm between the barbell and the body's centre of mass, increasing the torque that the lifter's muscle must counteract to keep balance. This increases the demand on the back extensors and reduces efficiency, as torque (τ = F ⋅ d), is maximised when d is larger. Keeping the bar closer minimises d, reducing torque and allowing for a more effective lift. High Jump 1. A lower run up velocity than for a long jump athlete: A lower run-up velocity in the high jump allows for better control during the takeoff phase, optimising vertical force application rather than horizontal. This slower approach aids in achieving a greater vertical impulse (Impulse = F ⋅ t), maximising lift-off height by directing more force vertically, which is crucial for clearing the bar. 2. Leaning backwards when planting the take-off foot just before take-off: Leaning backwards when planting the take-off foot increases the range of motion for forward rotation, allowing the athlete to generate greater angular momentum (L = Iω), where I is the moment of inertia and ω is angular velocity. This increased angular momentum contributes to elevating the body's centre of mass during take-off, enhancing jump height. The backward lean facilitates a more forceful hip and knee extension, maximising vertical take-off velocity and improving the effectiveness of the high jump. 3. Lowering the centre of gravity and extending the shoulders and elbows prior to take-off: Lowering the centre of gravity before take-off increases the distance and time over which force is applied against the ground, enhancing the impulse (Impulse = F ⋅ t) and resulting in a greater vertical take- off velocity. Extending the shoulders and elbows raises the body's centre of mass at take-off, effectively increasing the initial height (h) from which the jumper ascends, maximising potential energy (P E = ma g h). 4. Swinging the arms and free leg (non-take-off leg) up during take-off: Swinging the arms and free leg upward during take-off adds their mass and velocity to the body's upward motion, increasing the vertical momentum (p = mv). This action enhances the vertical ground reaction force, resulting in  a greater impulse (Impulse = F ⋅ t), which elevates the vertical take-off velocity. According to h = v2 2g , where h is jump height and g is gravitational acceleration, a higher take-off velocity leads to a more effective high jump. 5. Keeping the elbows flexed as the arms are swung up during take-off: Keeping the elbows flexed while swinging the arms upward reduces the arms' moment of inertia (I = mr 2 ) by decreasing the distance (r) from the shoulder joint to the mass of the arms. This reduction allows for a greater angular acceleration (α) for a given torque (τ = Iα), enabling the arms to swing upward more rapidly. The faster arm swing increases upward momentum, enhancing the vertical ground reaction force and resulting in a higher vertical take-off velocity, thereby improving the effectiveness of the high jump.