A ball is dropped from a height 16 m above the ground at t-0.0. Each time it bou

Question

A ball is dropped from a height 16 m above the ground at t-0.0. Each time it bounces from the ground, its rebound speed is 60% of its impact speed. At the instant the first ball hits the ground, a second ball is released from the same place. Take g = 9.8 m/s2. At what height above the ground will they collide? (Take the ground as y-0.0 m) Hint: Fig 1 describes the situation First ball A, becomes ball B after collision. Second ball C Ref Fig 1: Find VfA for first ball when YA-0.0 (V when it hits ground) VfA 0.6ViB for first ball as it goes upwards (call it ball B now) Ref Fig 2: Use Yf = Yi + ViT + (1/2)aT2 for both balls. (Vi for ball C = 0.0). Since Yf's are equal at collision point, set both equations equal to each other, i.e. Plug in values to get T (same for both at collision point) Use T to get Yf

Explanation / Answer

for the first ball, Vi=0

yo=16 m

y=0

Vf1 = velocity of first ball on hitting the ground

Vf12=Vo2+2 g(Y-Yo)

Vf12=0+2 (-9.8)(0-16)

Vf1=17.71m/s

velocity of ball after rebounding from the ground= 0.6*17.71=10.63 m/s

for first ball after rebounding from the ground

Yf= yi+Vit+1/2 gt2

Yf=10.63t+1/2 gt2 --------------(1)

for the second ball,

Vi=0, Yi=16

Yf= yi+Vit+1/2 gt2

Yf= 16+0(t)+1/2 gt2

Yf= 16+1/2 gt2 ----------------------(2)

equating the equation (1) and equation (2), we get

10.63t+1/2 gt2 =16+1/2 gt2

10.63t=16

t= 1.51 s

so from equation (2), we plugin the value of t

Yf=16+1/2 gt2

Yf=16+1/2(-9.8)(1.51)2

Yf=4.83m

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