KPE 263 Introductory Biomechanics Projectile Motion PDF

KPE 263 Introductory Biomechanics Week 3 Projectile motion Projectile Motion Bodies/Objects projected into the air are projectiles The object is moving through the air unassisted. The only forces acting on the object is the acceleration due to gravity and air resi...

KPE 263 Introductory Biomechanics Week 3 Projectile motion Projectile Motion Bodies/Objects projected into the air are projectiles The object is moving through the air unassisted. The only forces acting on the object is the acceleration due to gravity and air resistance. 01/20/25 2 Projectile Motion In the absence of air resistance flight path is a parabola 01/20/25 3 Projectile Motion Horizontal and Vertical Components The path of a projectile has a vertical AND a horizonal component. These components must be analyzed separately. The vertical component is influenced by gravity. 01/20/25 4 Projectile Motion Gravity The force of gravity acting on a projectile results in a constant vertical acceleration of the body Acceleration due to gravity = -9.81m/s2 01/20/25 5 Projectile Motion Horizontal and Vertical Components The path of a projectile has a vertical AND a horizonal component. These components must be analyzed separately. The vertical component is influenced by gravity. The distance travelled in the vertical component is the height of the projectile relative to the starting position. The horizontal component is not influenced by outside forces. The distance travelled in the horizonal is the distance travelled beyond the starting position. 01/20/25 6 Projectile Motion Horizontal and Vertical Components The path of a projectile has a vertical AND a horizonal component. These components must be analyzed separately. At the apex, velocity in the vertical component is 0 m/s. 01/20/25 7 Projectile Motion We assume that air resistance is negligible. Once in the air, the vertical component of velocity is always changing due to the influence of gravity (a = -9.81 m/s2). However, the horizontal component of velocity remains constant as horizontal acceleration is 0 m/s2. 01/20/25 8 Linear Acceleration Interpreting Linear Acceleration - Direction of Motion + Increasing Velocity in a= + Positive Direction Decreasing Velocity in a= - Positive Direction Increasing Velocity in a= - Negative Direction Decreasing Velocity in a= + 01/20/25 Negative Direction 9 Projectile Motion vy vx vy v vx Horizontal velocity (vx) of object remains constant (assuming negligible air resistance) Vertical velocity (vy) is changing (i.e., object is accelerating) at a constant rate of -9.81 m/s2 01/20/25 10 Projectile Motion Solving projectile motion problems (shotput example) Break up the problem into manageable parts 1. Vertical up vy=0 2. Vertical down vy =8.55m/s Vr =13.3m/s 3. Horizontal Θ =40° Vx = 10.2m/s H h =2.2m 01/20/25 x 11 Projectile Motion Solving projectile motion problems (shotput example) 1. Vertical up Given R.T.F Vi=8.55 m/s sf (H) = 5.92m vf=0m/s Vf = 0 m/s tf = 0.87s Si = 2.2m vi =8.55m/s a = -9.81m/s2 Vr =13.3m/s ti =0 𝑣𝑓2 = 𝑣𝑖2 + 2𝑎(𝑠𝑓 − 𝑠𝑖 ) 𝑣𝑓 = 𝑣𝑖 + 𝑎∆𝑡 Θ =40° 0 = 8.552 + 2(−9.81)(𝑠𝑓 − 2.2) 0 = 8.55 + (−9.81)(tf − 0) -73.10 = −19.62(𝑠𝑓 − 2.2) −8.55 = (−9.81)tf H -73.10 = −19.62𝑠𝑓 + 43.16 tf = 0.87𝑠 si =2.2m -116.26 = −19.62𝑠𝑓 𝑠𝑓 = 5.92𝑚 01/20/25 12 Projectile Motion Solving projectile motion problems (shotput example) Given R.T.F 2. Vertical down Vi=0 m/s tf = 1.96s Si = 5.92 Vf = −10.78 𝑚/𝑠 vi=0 Sf = 0 ti=0.87 a = -9.81m/s2 ti =0.87 𝑣𝑓2 = 𝑣𝑖2 + 2𝑎(𝑠𝑓 − 𝑠𝑖 ) 𝑣𝑓 = 𝑣𝑖 + 𝑎∆𝑡 𝑣𝑓2 = 0 + 2 (−9.81)(0 − 5.92) −10.78 = 0 + (−9.81)(𝑡𝑓 − 0.87) 𝑣𝑓2 = (-19.62)(-5.92) −10.78 = −9.81𝑡𝑓 + 8.53 𝑣𝑓 = 116.27 −19.31 = −9.81𝑡𝑓 H (si) = 5.92 𝑣𝑓 = −10.78 𝑡𝑓 = 1.96𝑠 01/20/25 x 13 Projectile Motion Horizontal Component The range (horizontal displacement) of a projectile can be calculated using a formula derived from our equations of motion 𝑣𝑟2 ∙ 𝑠𝑖𝑛𝜃 ∙ 𝑐𝑜𝑠𝜃 + 𝑣𝑥 ∙ 𝑣𝑦2 + 2𝑔ℎ 𝑅𝑎𝑛𝑔𝑒 𝑅 = 𝑔 vy vr vx h R 01/20/25 14 Projectile Motion Solving projectile motion problems (shotput example) 3. Horizontal 𝑣𝑟2 ∙ 𝑠𝑖𝑛𝜃 ∙ 𝑐𝑜𝑠𝜃 + 𝑣𝑥 ∙ 𝑣𝑦2 + 2𝑔ℎ 𝑅𝑎𝑛𝑔𝑒 𝑅 = 𝑔 vy=0 𝑅𝑎𝑛𝑔𝑒 𝑅 = 20.08m vy =8.55 Vr =13.3 Θ =40° Vx = 10.2 H h =2.2 x 01/20/25 15 Projectile Motion Solving projectile motion problems (shotput example) Therefore the shotput travels a total of 20.08m horizontally, reaches a peak height of 5.92 m above the ground, lands with a velocity of 10.78m/s and is in flight for a total of 1.96s vy=0 vy =8.55 Vr =13.3 Θ =40° Vx = 10.2 H h =2.2 x 01/20/25 16 Projectile Motion Example: A soccer ball is kicked with an initial resultant velocity of 5 m/s at a projection angle of 25°. Assuming that projection and landing heights are the same and neglecting air resistance, identify the following quantities: a) The ball’s horizontal velocity 0.5 s into flight b) The ball’s horizontal velocity midway through flight c) The ball’s horizontal velocity immediately before ground contact d) The ball’s vertical velocity at apex of flight e) The ball’s vertical velocity midway through flight f) The ball’s vertical velocity immediately before ground contact 01/20/25 17 01/20/25 18 01/20/25 19 01/20/25 20 Projectile Motion A projectile’s trajectory is influenced by: Gravity Air resistance Projection parameters Projection velocity Projection angle Projection height 01/20/25 21 Projectile Motion Influence of Projection Velocity Resultant vector of horizontal and vertical velocity components Influences maximum height (apex) and maximum distance (range) Projection height and angle held constant in this example 01/20/25 22 Projectile Motion Projection Angles Influences relative magnitude of horizontal and vertical velocities 01/20/25 23 Projectile Motion Projection Angles Influences relative magnitude of horizontal and vertical velocities Projection height and velocity 01/20/25 held constant 24 Projectile Motion Projection Height Relative projection height (RPH) = vertical take-off position – vertical landing position: Takeoff = landing; RPH = 0 Takeoff > landing; RPH >0 Takeoff < landing; RPH< 0 01/20/25 25 Projectile Motion Projection Angles Influences relative magnitude of horizontal and vertical velocities Projection height and velocity 01/20/25 held constant 26 Projectile Motion Projectile Motion can be analysed using our constant acceleration equations. 1 𝑣𝑓 = 𝑣𝑖 + 𝑎∆𝑡 𝑠𝑓 = 𝑠𝑖 + 𝑣𝑖 ∆𝑡 + 𝑎∆𝑡 2 2 1 𝑠𝑓 = 𝑠𝑖 + (𝑣𝑓 + 𝑣𝑖 )∆𝑡 𝑣𝑓2 = 𝑣𝑖2 + 2𝑎(𝑠𝑓 − 𝑠𝑖 ) 2 For projectiles: Vertical component is affected by the acceleration due to gravity (-9.81 m/s2). Horizontal component is not affected by external force, so acceleration will be 0 m/s2. 01/20/25 27 A field goal kicker can kick the ball off the ground at 30m/s at an angle of 30° relative to the positive horizontal. If the upright cross- bar is 3m off the ground, from how far away can they kick a field goal? 01/20/25 28 01/20/25 29 01/20/25 30 01/20/25 31 01/20/25 32 Summary Displacement for each component Slope 𝑟𝑖𝑠𝑒 ∆𝑥 ∆𝑥 = 𝑥2 − 𝑥1 𝑆𝑙𝑜𝑝𝑒 = 𝑆𝑙𝑜𝑝𝑒 = 𝑟𝑢𝑛 ∆𝑡 ∆𝑦 = 𝑦2 − 𝑦1 Speed First Central Difference Method 𝐷𝑖𝑠𝑡𝑎𝑐𝑒 𝑆𝑝𝑒𝑒𝑑 = 𝑡𝑖𝑚𝑒 𝑖 + 1 − (𝑖 − 1) 𝑣𝑖 = Velocity 𝑡𝑖+1 − 𝑡𝑖−1 ∆𝑠 𝑣= ∆𝑡 Velocity can also be estimated graphically 𝑥2 − 𝑥1 𝑦2 − 𝑦1 using the slopes and the local extremes 𝑣𝑥 = 𝑣𝑦 = 𝑡2 − 𝑡1 𝑡2 − 𝑡1 01/20/25 33 Summary Constant Acceleration Equations First Central Difference Method 𝑣𝑓 = 𝑣𝑖 + 𝑎∆𝑡 𝑖 + 1 − (𝑖 − 1) 𝑣𝑖 = 1 𝑡𝑖+1 − 𝑡𝑖−1 𝑠𝑓 = 𝑠𝑖 + 𝑣𝑖 ∆𝑡 + 𝑎∆𝑡 2 2 𝑣𝑓2 = 𝑣𝑖2 + 2𝑎(𝑠𝑓 − 𝑠𝑖 ) Acceleration can also be estimated graphically using the slopes and the local extremes of the velocity-time curve 1 𝑠𝑓 = 𝑠𝑖 + (𝑣𝑓 + 𝑣𝑖 )∆𝑡 2 𝑣𝑟2 ∙ 𝑠𝑖𝑛𝜃 ∙ 𝑐𝑜𝑠𝜃 + 𝑣𝑥 ∙ 𝑣𝑦2 + 2𝑔ℎ 𝑅𝑎𝑛𝑔𝑒 𝑅 = 𝑔 01/20/25 34 Up Next Angular Kinematics 5th edition: Chapter 10 pg 390-396 and 399-406 4th edition: Chapter 9 pg 319-324 and 327-334 01/20/25 35