Question

A ball is projected from the ground at a speed of $10m{s}^{-1}$ making an angle ${30}^{\circ }$ with the horizontal. Another ball is simultaneously released from a point on the vertical line along the maximum height of the projectile. Both the balls collide at the maximum height of the projectile. What was the initial height of the freely falling body?( Assume $g=10m{s}^{-2}$)

• A. $5m$
• B. $10m$
• C. $4m$
• D. $2.5m$

Answer 1

The correct option is D $2.5m$

Given
initial velocity $u=10m/s$
Angle of projection $\theta ={30}^{\circ }$
Time of flight is
$T=\frac{2usin\theta }{g}=\frac{2\times 10\times \frac{1}{2}}{10}$
$T=1s$
Max height is,
$H=\frac{(usin\theta {)}^2}{2g}=\frac{{10}^2\times \frac{1}{4}}{20}$
$H=\frac{5}{4}m$
Time taken by projectile to reach maximum height is $T/2$.
The ball released travels $h-H$ in time interval of $T/2$.
So,
$h-H=\frac{1}{2}g{\left(\frac{T}{2}\right)}^2$ $(\because g=10m{s}^{-2})$
$h-\frac{5}{4}=\frac{5}{4}$
$h=\frac{10}{4}$
$h=2.5m$