Question

A body is projected horizontally from the top of a tower with a velocity of $10m/s$ . If it hits the ground at an angle of ${45}^o$ , the vertical component of velocity when it hits ground (in $m/s$) is:

• A. $10\sqrt{2}$
• B. $5\sqrt{2}$
• C. $5$
• D. $10$

Answer 1

The correct option is D $10$

Let the body hit the ground with velocity $v$ and its vertical componant of velocity be $v^{\prime }$.

As no external force acts on the body in horizontal direction, thus the horizontal velocity of the body remains the same i.e $v^{\prime \prime }=10$ $m/s$

$\therefore$ $vsin{45}^o=v^{\prime }$ ............(1)

Also $vcos{45}^o=10$ ..........(2)

Dividing (1) and (2) we get, $\frac{v^{\prime }}{10}=tan{45}^o=1$

$⟹v^{\prime }=10$ $m/s$