Question

A projectile thrown with velocity $v$ making angle $\theta$ with vertical gains maximum height $H$ in the time for which the projectile remains in air, the time period is

• A. $\sqrt{\frac{H\sin \theta }{g}}$
• B. $\sqrt{\frac{H\cos \theta }{g}}$
• C. $\sqrt{\frac{4H}{g}}$
• D. $\sqrt{\frac{8H}{g}}$

Answer 1

The correct option is D $\sqrt{\frac{8H}{g}}$

Maximum height, $H=\frac{v^2{\sin }^2(90-\theta )}{2g}......\left(i\right)$
Time of flight, $T=\frac{2v\sin (90-\theta )}{g}......\left(ii\right)$

From $\left(i\right),\frac{v\cos \theta }{g}=\sqrt{\frac{2H}{g}}$
From $\left(ii\right),$
$T=2\sqrt{\frac{2H}{g}}=\sqrt{\frac{8H}{g}}$