Question

A stone thrown vertically passes a certain point P at the end of 2 seconds and 8 seconds respectively. Find the maximum height reached by the stone .$(Takeg={10ms}^{-2})$

• A. 625 m
• B. 125 m
• C. 225 m
• D. 350 m

Answer 1

The correct option is B 125 m

Let initial velocity $u$ and point 'P' is at height $h$.

At the end of 2 seconds,

Using $s=ut+\frac{1}{2}a{t}^2$ $\Rightarrow h=u\times 2-\frac{1}{2}g\times 2^2$ $\Rightarrow h=2u-20$ .......1

At the end of 8 seconds,

$\Rightarrow h=u\times 8-\frac{1}{2}g\times 8^2$ $\Rightarrow h=8u-320$ ....2

Solving equations 1 and 2, $\Rightarrow u=50m/s$,

At maximum height , velocity $v=0$,

Using $v^2=u^2+2as$ $\Rightarrow 0={50}^2-2\times 10\times s$ $\Rightarrow s=125m$,

$\therefore$ Maximum height $=125m$