Question

A ball is thrown upward with a certain speed.It passes through the same point at 3 second and 7 second from the start. the maximum height achieved by ball is

Answer 1

Given Data:

Time taken to repeat same position, $t_1=3sec,t_2=5sec$

As we know the ball is thrown upward and after crossing point at 3 seconds it will reach at the top most position within,$=\frac{7-3}{2}=2sec$ [ because time of ascent and descent from this point are same]

Therefore the ball will take 3+2 or 5 sec to reach at the top from the bottom.

Therefore for downward direction the motion will be;

Initial velocity, $u=0m{s}^{-1},g=10{ms}^{-2},t=5sec$

So height will be, $h=ut+\frac{1}{2}{gt}^2⟹h=0+\frac{1}{2}\left(10\right)\times 5^2=125m$