Question

When a projectile is thrown at angle $\theta$ with the vertical it reaches the maximum height D. The time of flight of the projectile is

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

Answer 1

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

Lets consider, $u=$ initial velocity of the projectile

$\theta =$ angle of projection

Maximum height, $D=\frac{u^{2}si{n}^2\theta }{2g}$

$u^{2}si{n}^2\theta =2gD$

$usin\theta =\sqrt{2gD}$. . . . . . .(1)

Time of flight of the projectile,

$T=\frac{2usin\theta }{g}$

$T=2\frac{\sqrt{2gD}}{g}$

$T=\sqrt{\frac{8gD}{g^2}}$

$T=\sqrt{\frac{8D}{g}}$

The correct option is D.