Question

A particle is projected at an angle $\theta$ to the horizontal and it attains a maximum height H. The time taken by the projectile to the highest point, of its path is

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

Answer 1

The correct option is B $\sqrt{\frac{2H}{g}}$

$V_y=U_y-gt$
$t=\frac{U_y}{g}$
$H=U_{y}t-\frac{1}{2}g{t}^2$
$H=gt\times t-\frac{1}{2}g{t}^2$
$H=\frac{1}{2}g{t}^2$
$t=\sqrt{\frac{2H}{g}}$