Question

A stone is thrown vertically upward with an initial velocity of $40m/s$. Taking $g=10m/s^2$, find the maximum height reached by the stone. What is the net displacement and the total distance covered by the stone?

Answer 1

Given : Initial velocity of stone $u=40$ $m/s$

Let the maximum height reached and time taken to reach that height be $H$ and $t$ respectively.

Velocity of the stone at maximum height is zero i.e $v=0$

Using, $v^2-u^2=2aH$ where $a=-g=-10$ $m/s^2$

$0-(40{)}^2=2(-10)H$ $⟹H=80$ m

The stone, after reaching the maximum height, starts to fall down and reach the surface again.

$\therefore$ Net displacement of the stone is Zero

Total distance covered by the stone $S=2H=2\times 80=160$ m