Q. A ball is dropped from a height d above the ground. It hits the ground and bounces up vertically to a height d/2. How will its vel. v vary with height h above the ground. (Neglect subsequent motion and air resistance)

Please tell apt graph for above question & explain it.

before hitting ground for first time, v=√(2g(d-h))

obviously, coefficient of restitution=1/√2,

let answer be 'v'

g(d-h)-v²=2gh

i can't suggest a graph , but as you may see the coefficient of restitution 1/sqrt(2) . so if the subsequent motion is considered it will keep on deceresing next time it will be 1/4 ,1/8 and so on....... the ball will finally come to rest . between two collisions the acceleration will b const. (g)

Why have u taken vel. before hitting the ground for the 1st time as v=[(sqrt)2g(d-h)], shouldn't it be v=[(sqrt)2gd] only.

And second thing what is meant by coefficient of restitution?

velocity of the ball before hitting ground=sqrt(2gd)

& the coeffient of restitution=1/sqrt(2)

so its velocity will vary with height by 1/2,1/4,1/8,1/16................ so on

& finally ball comes to rest.