Question

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 h_0 = 1.5 m with an initial speed of V_0, = 28 m/s The coefficient of restitution between the golf ball and the concrete is e = 0.92. Determine the maximum vertical height the golf ball will reach after its third bounce. h_3. Determine a formula for the maximum vertical height the golfball will reach after its nth bounce, h_n, interms of the variables, h_0, g, e and n. 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?

Answer 1

consider maximum height at a certain bounce number be h.
at maximum height vertical speed =0

time taken to reach ground=sqrt(2*h/g)

velocity in this time=initial velocity+acceleration*time

=0+g*sqrt(2*h/g)=sqrt(2*g*h)

as coefficient of restitution is 0.92,

then the velocity with which the ball will bounce back=0.92*sqrt(2*g*h)

then maximum height reached in next stage=initial velocity^2/(2*g)

=0.92^2*2*g*h/(2*g)=0.92^2*h

using the same analysis,

for nth bounce, maximum height=0.92^(2*n)*h0

part a:

using n=3, h3=0.92^(6)*1.5=0.90953 m

part b:

maximum vertical height after nth bounce=e^(2*n)*h0

part c:

let at nth bounce, bounce height=0.5*42.7 mm=21.35 mm=0.02135 m

then 0.92^(2*n)*1.5=0.02135

==>0.92^(2*n)=0.014233

taking logarith,

2*n=50.996

==>n=25.498

hence after 26th bounce, the ball will start rolling.