Question

A projectile thrown with velocity $v$ making angle ${30}^{\circ }$ with vertical gains maximum height $100m$ in the time for which the projectile remains in air, the time of flight of the projectile in seconds is (take $g=10{m/s}^2$)

• A. $\sqrt{120}$
• B. $\sqrt{40}$
• C. $\sqrt{60}$
• D. $\sqrt{80}$

Answer 1

The correct option is D $\sqrt{80}$

Given $\theta ={30}^{\circ }$
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 $\text{(i)}$, $\frac{v\sin (90-\theta )}{g}=\sqrt{\frac{2H}{g}}$
$\Rightarrow T=2\sqrt{\frac{2H}{g}}=\sqrt{\frac{8\times 100}{10}}=\sqrt{80}$
Hence, the correct answer is option (d)