Question

A boy standing below a tree throws a ball vertically upward with initial speed u m/s. the ball gets trapped into leaves for 2 sec and then returns into the boy's hands. if the height at which ball was trapped was 20 m and the total time after which ball again returned back to boy's hand is 5 sec, then calculate the initial speed u?

Answer 1

Given,
Height at which ball was trapped, h=20 m
Total time after which ball again returned back to hand=5 sec
Time to reach maximum height,t=5/2=2.5 s
Since ball gets trapped into leaves for 2s.
So, t=2.5-2=0.5 s
$$\begin{gathered} {\mathrm{v}}^2-{\mathrm{u}}^2=2\mathrm{gh} \\ 0-{\mathrm{u}}^2=2\times (-10)\times 20=-400 \\ \mathrm{u}=20\mathrm{m}/\mathrm{s} \end{gathered}$$