GP1 Lesson 05 Kinematics in 2D PDF

 Lesson 05 Kinematics in Two-Dimensions John Myles D. Canuto, EcT, LPT OBJECTIVES At the end of the lesson, the students should be able to: 1. Explain with equations and diagrams the horizontal and vertical motion of a projectile launched at various angles; 2. Determine the position...

 Lesson 05 Kinematics in Two-Dimensions John Myles D. Canuto, EcT, LPT OBJECTIVES At the end of the lesson, the students should be able to: 1. Explain with equations and diagrams the horizontal and vertical motion of a projectile launched at various angles; 2. Determine the position and velocity of a projectile when its initial velocity and position are given; and 3. Determine the range, the maximum height, Projectile Motion The two-dimensional motion of a thrown object, influenced by the pull of gravity, is called projectile motion. Its path is called Regarding this motion, air resistance will not be considered. A parabolic trajectory will be formed if air resistance is neglected. A projectile will accelerate downward due to the pull of gravity downward. The acceleration along the y-axis will point to the direction toward the ground. Hence, the value of the acceleration will be negative and equal to the acceleration caused by the pull of gravity on Earth (g). This acceleration is equal to -9.8 m/s2. However, along the x-axis, no acceleration occurs since the pull of gravity is directed downward. Hence, the Components of the velocity moving im projectile motion: v0X = v0cosӨ v0Y = v0sinӨ At the highest point along the projectile, the velocity along the y-axis will be 0.0 m/s. Once the object reaches this point, it stops moving upward and begins to go downward. As the object continues to move, the  time of flight’. Once the object in projectile motion reaches its maximum height, it will proceed to go downward until it reaches the ground. The half-time signifies that the object has reached half of the total range the trajectory could reach. HALF TIME OF FLIGHT vf = vo + at vfy = voy + ayt t= To obtain the value of the maximum height, we have the formula: dy = voyt + ½ ayt 2  The range of a trajectory is the total horizontal distance of an object moving in projectile motion. The determination of the total range is dependent on the velocity along Range All throughout the projectile, the value of the x-velocity does not change. Thus, the final velocity is also the initial velocity. Start with the displacement formula: d = vot + ½ at2 Since it is along the x-direction, subscript this equation for x: dx = voxt + ½ axt2 However, no acceleration is present along that direction, the equation becomes dx = voxt velocity is the same all throughout the axis, the projectile will maintain its initial x-velocity all throughout the trajectory.  Formula for Displacement x = v0xt (horizontal) y = ½ gt (vertical) 2 Formula for Velocity vx = v0x (horizontal) vy = v0y + gt (vertical) SAMPLE PROBLEMS Problem 1 A skier leaves the jump horizontally with an initial velocity of 25 m/s. The initial height at the end of the ramp is 80 m above the landing position. (a) How long is the skier in the air? (b) How far does the skier travel horizontally? (c) What are the horizontal and vertical components of a. The time spent in the air is a function of only vertical parameters. Given: v0y = 0, a = +9.8 m/s2, y = +80 m Find: t =? We need an equation that contains t and not vf. y = v0yt + ½ gt2 Setting v0y = 0 and solving for t, we obtain y = ½ gt2 and t = t = = 4.04 s It takes a time of 4.04 s for the skier to reach the landing point. b. Since the horizontal velocity is constant, the range is determined only by the time of flight. x = v0xt = (25 m/s)(4.04 s) c. The horizontal component of the velocity is unchanged and, therefore, equal to 25 m/s at the landing point. The final vertical component is given by Vy = gt = (9.8 m/s2)(4.04 s) = 39.6 m/s Problem 2 A projectile is fired with an initial velocity of 80 m/s at an angle of 30 degrees above the horizontal. Find (a) its position and velocity after 6 s, (b) the time required to reach the maximum height, and (c) the a. The horizontal and vertical components of the initial velocity are v0x = v0 cos Ө = (80 m/s)(cos 30) = 69.3 m/s v0y = v0 sin Ө = (80 m/s)(sin 30) = 40.0 m/s The x-component of its position after 6 s is given by x = v0xt = (69.3 m/s)(6 s) = 416 m The y-component of its position at this time is y = v0yt + ½ gt2 y = (40 m/s)(6 s) + ½ (-9.8 m/s2)(6 s)2 y = 240 m – 176 m = 64.0 m The position after 6 s is 416 m downrange and In computing its velocity at this point, we first recognize that the x component of the velocity does not change. Thus, vx = v0x = 69.3 m/s The y-component of velocity must be found from vy = v0y + gt vy = 40 m/s + (-9.8 m/s2)(6 s) vy = 40 m/s – 58.8 m/s The magnitude of the velocity is v= v = 71.8 m/s The angle is found from the tangent function tan Φ = Abs = 0.271 Angle = 15.2 degrees S of E b. At the maximum point in the path, the y-component of the velocity is zero. Hence, the time to reach the maximum height can be found from vy = v0y + gt where vy = 0 0 = (40 m/s) + (-9.8 m/s2)t t = (40 m/s) / (9.8 m/s2) t = 4.08 s dy = (40 m/s)(4.08 s) + ½ (-9.8 m/s2) 2 c. The range of the projectile can be found by recognizing that the total time t’ of the entire flight is equal to twice the time to reach the highest point. Therefore, t’ = 2(4.08 s) = 8.16 s The range is R = v0xt’ = (69.3 m/s)(8.16 s) Problem 3 In a movie production, a stunt person must leap from a balcony of one building to a balcony 3.0 m lower on another building. If the building are 2.0 m apart, what is the minimum horizontal velocity the stunt person must have to accomplish the jump? Assume no air resistance and that ay Given: Δy = -3.0 m, Δx = 2.0 m Δy = ½ ay(Δt)2 Δt = = vx = vavg, x = = x = (2.0 m) = 2.6 m/s Thanks! Do you have any questions? [email protected] du.ph CREDITS: This presentation template was created by Slidesgo, and includes icons by Flaticon, and infographics & images by Freepik