Question

A ball falls from a height of 1m, hits the ground and rebounds with half itsvelocity just before impact. then after rising it falls and hits the ground and againrebounds with half its velocity just before impact, and so on. the total distancetraveled by the ball till it comes to rest on the ground is

Answer 1

A ball falls from a height of 1 m, hits the ground, and rebounds with half its velocity just before impact. After rising it falls and hits the ground and again rebounds with half its velocity just before impact and so on. The total distance traveled by ball till it comes to rest is 2m

Answer

2m

Answer 2

Given info : A ball falls from a height of 1m, hits the ground and rebounds with half its velocity just before impact. then after rising it falls and hits the ground and again rebounds with half its velocity just before impact, and so on.

To find : The distance travelled by the ball till it comes to rest on the ground is...

solution : velocity of ball before striking the ground , v = -√(2gh) = -√(2g)

now ball strikes and rebounds with half of its velocity. I.e., velocity of ball now = √(2g)/2 = √(g/2)

so height reached by ball = u²/2g = {√(g/2)}²/2g = 1/4

again falls from this height and strikes grounds with half of its velocity.

velocity of ball now = √(g/2)/2 = √(g/8)

so height reached by ball = (g/8)/2g = 1/16

similarly in 3rd time height reached by ball = 1/64

and so on...

total distance = 1 + 2(1/4 + 1/16 + 1/64 + ...)

here we multiplied by 2 because ball reaches some height then falls from that height. so total distance = 2 × height reached by ball.

now using GP formula,

1/4 + 1/16 + 1/64 + ... = (1/4)/(1 - 1/4) [ S = a/(1 - r) ]

= 1/3

so total distance = 1 + 2/3 = 5/3

Therefore the distance traveled by the ball till it comes to rest is 5/3 m