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 of flight of the projectile 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 $\sqrt{\frac{8H}{g}}$


Angle of projection of the projectile with the horizontal = $90-$$\theta$
Maximum height, $H=\frac{v^2{\sin }^2(90-\theta )}{2g}......\text{(i)}$
Time of flight, $T=\frac{2v\sin (90-\theta )}{g}......\text{(ii)}$
From $\text{(i)}$, $\frac{v\sin (90-\theta )}{g}=\sqrt{\frac{2H}{g}}$
$\Rightarrow T=2\sqrt{\frac{2H}{g}}=\sqrt{\frac{8H}{g}}$
Hence, the correct answer is option (d)