Answer to Question #134060 in Mechanics | Relativity for David P. Costello

Question

A tennis ball is dropped from 1.21 m above the ground. It rebounds to a height of 1.05 m. With what velocity does it hit the ground? The acceleration of gravity is 9.8 m/s 2 . (Let down be negative.) Answer in units of m/s.

With what velocity does it leave the ground? Answer in units of m/s

Answer 1

solution

given data

initial velocity when dropped(u)=0 m/s

distance traveled(s)=1.21 m/s

distance traveled after leave the ground(s₁)=1.05 ms

according to third law of motion

$v^2=u^2+2gs$

initial velocity u=0

$v^2=2gs\space or\space v=\sqrt{2gs}$

by putting the value of s

$v_{hitting}=\sqrt{2\times9.8\times1.21}=4.86m/s$

after leaving ground at 1.21 m final velocity will be zero

so velocity just after left the ground will be

$v_{2}=\sqrt{2\times9.8\times1.05}=4.53m/s$

so velocity of hitting the ground is 4.86 m/s and after hitting is 4.53 m/s .