Question

A stone is thrown from the ground just crosses a wall of height $5$ m in $2$ s. If the wall is at the horizontal distance 25 m, from the point of projection of the stone, find at what distance the stone falls behind the wall.$(Takeg=10{ms}^{-2})$

• A. $31.25$ m
  
• B. $6.25$ m
• C. $25$ m
• D. $32.5$ m

Answer 1

The correct option is B $6.25$ m

Let velocity of projectile is $u\hat{i}+v\hat{j}$.

Given , in time 2 s, Horizontal distance= 20 $m$, Vertical Distance = 5 $m$,

Using $s=v\times t$, $\Rightarrow 20=u\times 2$ $\Rightarrow u=12.5m{s}^{-1}$

Using $s=ut-\frac{1}{2}g{t}^2$ $\Rightarrow 5=v\times 2-\frac{1}{2}g\times 2^2$ $\Rightarrow v=12.5m{s}^{-1}$,

At maximum height , Vertical velocity= 0 , Time taken by stone to reach at maximum height is $t$.

$\Rightarrow 0=v-gt$ $\Rightarrow t=1.25s$

$\therefore \text{Time of Flight}=2\times t=2.5s$,

Range $R=u\times \text{Time of Flight}$ $\Rightarrow R=12.5\times 2.5=31.25m$,

So the distance between the wall and final point of stone = 31.25-25= 6.25 $m$.