Question

A ball is thrown vertically upward returns to the thrower in 4 sec. with what velocity did the thrower threw the ball.

Answer 1

The ball travels twice the length of the maximum height in 6 seconds i.e., it takes 3 seconds to go up and 3 seconds to come down.

At the topmost point the velocity is zero and whatever kinetic energy was provided to the ball initially transforms into potential energy when it comes down the opposite happens.

Using the kinematic equation-

v=u+atv=u+at

substituting v=0 m/s ,u =u (we don’t know it yet),a=-g (when the ball is going from bottom to top) and t=3 s:

0=u−3∗g0=u−3∗g

therefore,u=3g=3∗9.8=29.4m/stherefore,u=3g=3∗9.8=29.4m/s

For the maximum height ,use the conservation of energy i.e.,

1/2∗m∗v2=m∗g∗H1/2∗m∗v2=m∗g∗H

The above equation becomes,

v2/2=g∗Hv2/2=g∗H

H=max height.

H=v2/(2∗g)=(29.4)2/(2∗9.8)=44.1mH=v2/(2∗g)=(29.4)2/(2∗9.8)=44.1m

Initial Velocity , u=29.4 m/s

Max height, H=44.1 m

:)