Question

A projectile is projected from ground with velocity $v_0$ at an angle $\theta$ with horizontal. What is the velocity of projectile when its vertical displacement is equal to half of the maximum height attained?

• A. $\sqrt{3}{v}_{0}cos\theta$
• B. $\frac{v_0}{\sqrt{2}}sin\theta$
• C. $\frac{v_0}{\sqrt{2}}cos\theta$
• D. $\sqrt{5}{v}_0$

Answer 1

The correct option is B $\frac{v_0}{\sqrt{2}}sin\theta$

Initial vertical component of the velocity, $u=v_{0}sina$

Maximum height is given by, $v^2-u^2=-2gh$

At top most height, $v=0$

Then,

$0-(v_{0}sina{)}^2=-2gh$

$h=-\frac{(v_{0}sina{)}^2}{2g}$

$h^{\prime }=\frac{h}{2}$

$h^{\prime }=\frac{(v_{0}sina{)}^2}{4g}$

Vertical component of the velocity at this point

$v^{\prime 2}-u^2=-2g{h}^{\prime }$

$v^{\prime }=\sqrt{(v_{0}sina{)}^2-2g\frac{(v_{0}sina{)}^2}{4g}}$

$v^{\prime }=\frac{v_{0}sina}{\sqrt{2}}$