Projectile Motionand Impact: Just for fun, a golfer throws a golf ball horizonta
ID: 1790109 • Letter: P
Question
Projectile Motionand 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 h0= 1.5 m with an initial speed of V0= 15m/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, h0, 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?
Projectile Motion 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 = 15 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, h. (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 roling begin? hiExplanation / Answer
vh ^ 2 = 2gs
v^2 = ( vo ^2 + vh^2) = ( 15 ^ 2 + 2 * 9.8 * 1.5)
So v = 15.94 just as you have calculated.
But this figure is not useful within the problem.
What you have is that after the first bounce the ball will rise to 1.5 * 0.92
after the second it is ( 1.5 * 092) * 0.92
and after the third it is ( (1.5 * 0.92) * 0.92) * 0.92
in other words it is hn = 1.5 * 0.92 ^ n
For the last one you have hn < 0.021 m
hence 1.5 * 0.92^n < 0.021
0.92^n < 0.021/ 1.5
Which you could solve by "trial and error"
UPDATE to respond to your comment. The coefficient of restitution is the fraction of the height regained after any and every bounce by definition. ( it is actually can also a measure of the ENERGY recovered but that has pitfalls in practice).
If you had calculated the height after the first bounce ( 1.5 * 0.92 = 1.38 m) then the ball reaches 1.38 m
It then falls from 1.38 m to the ground . Now by definition after this bounce it must get back 1.38 * 0.92 = 1.27 m approx.
The reason that your speed is not useful is that it includes the horizontal speed. The coefficient of restitution only applies to the component of velocity perpendicular to the surface. The coefficient parallel to the surface does not alter on any bounce. So the coefficient only applies to a component of your velocity and not to the whole lot.
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