Question

Projectile Motion, Energy, and Impact: Just for fun, a golfer throws a golf ball horizontally through the air and watches it bounce again and again down a long straight concrete path. The ball is thrown horizontally from a height of ho = 1.5 m with an initial speed of Vo = 28 m/s. The coefficient of restitution between the golf ball and the concrete is e=0.92. (A) Determine the maximum vertical height the golf ball will reach after its third bounce, h3. (B) Determine a formula for the maximum vertical height the golf ball will reach after its nth bounce, hn, in terms of the variables, ho, g, e, and n. (C) The bouncing of a small sphere is modeled as dynamic particle motion. Particles are assumed to be point masses that take up no space, even though the diameter of a golf ball is about 42.7 mm in reality. Eventually, after enough bounces have occurred, the golf ball will cease bouncing and roll along the ground. If we suppose that rolling begins when the bounce height drops to half the sphere's true diameter, after how many bounces, m, will rolling begin? o

Answer 1

Part A: Let's use the conservation of energy to calculate the vertical comment of the velocity when the ball reaches the ground: E. = EF mgh+= mv? = m(V² + v2) Then v= 2gh The velocity after the first impact: vi = ev After the second impact: v2 = evi = ev After the third impact: V3 = e’v = e/2gh The maximum height reached is: hz = 3 = eh = (0.92) (1.5) = 0.91 m 29 Part B: For the nth bouncing ball: nmax = 29 Where: Vm = e" y = en 2gh Then: hmax = einh d Part C: In this case: hmax = = = 21.35 x 10-3m Then: e2nh = 21.35 x 10-3 0.922n (1.5) = 21.35 x 10-3 Solving for n: n = 25.49