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 reach the highest point, of its path is
- A. $\sqrt{H/g}$
- B. $\sqrt{2H/g}$
- C. $\frac{\sqrt{2H \sin \theta}}{g}$
- D. $\frac{\sqrt{2H}}{\sin \theta}$
Answer: B. $\sqrt{2H/g}$
Explanation: Max height $H = (u \sin \theta)^2 / 2g$. Vertical component $u_y = u \sin \theta = \sqrt{2gH}$. Time to max height $t = u_y / g = \sqrt{2gH}/g = \sqrt{2H/g}$.