Question

A projectile is thrown at angle b with vertical. It reaches a maximum height H. The time taken to reach highest point of its path is –

• A. $\sqrt{\frac{H}{g}}$
• B. $\sqrt{\frac{2H}{g}}$
• C. $\sqrt{\frac{H}{2g}}$
• D. $\sqrt{\frac{2H}{g\cos\beta}}$

Answer 1

Answer: (B)

$\sqrt{\frac{2H}{g}}$

Answer 2

H = $\frac{u^{2}\cos^{2}\beta}{2g}$ ̃ u cos b = $\sqrt{2gH}$

t = $\frac{u\cos\beta}{g}$= $\sqrt{\frac{2H}{g}}$