Question

A ball is thrown upward. It leaves the hand with a velocity of 15.5 m/s, having been accelerated through a distance of 0.480 m. Compute the ball's upward acceleration, assuming it to be constant.

Answer 1

there will be two case to solve this

case 1 if ball stops after travelling 0.480 m i.e. v=0m/s

initial velocity(u)= 15.5 m/s

acceleration= -a m/s^2

distance=S=0.480 m

by newton formula

v^2=u^2+2aS

a=u^2/2S=250.26 m/s^2

upward acceleration is -250.26 m/s^2

case 2 :If ball does not stop after travelling 0.480 m

then information is insufficient to calculate upward accleration and we will assume upward acceleration to be due to acceleration due to gravity i.e. upward acceleration = -9.8 m/s^2