Question

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

• A. $\sqrt{\frac{H\cos \theta }{g}}$
• B. $\sqrt{\frac{2H\cos \theta }{g}}$
• C. $\sqrt{\frac{4H}{g}}$
• D. $\sqrt{\frac{8H}{g}}$

Answer 1

Answer: (D)

$\sqrt{\frac{8H}{g}}$

replace $\theta$ by $90-\theta$ in above concepts, because it's angle from verticle here
From $\left(i\right)\frac{v\cos \theta }{g}=\sqrt{\frac{2H}{g}}$

From $\left(ii\right)$ $T=2\sqrt{\frac{2H}{g}}=\sqrt{\frac{8H}{g}}$