Q. A projectile is thrown at angle $\beta$ with vertical. It reaches a maximum height H. The time taken to reach the highest point of its path is
A) $\sqrt{\frac{H}{g}}$
B) $\sqrt{\frac{2H}{g}}$
C) $\sqrt{\frac{H}{2g}}$
D) $\sqrt{\frac{H}{g\cos\beta}}$
Explanation
(b) $Maximum\ height\ H = \frac{v^{2}\cos^{2}\beta}{2g}$
$or\ \ vcos\beta = \sqrt{2gH}$
$t = \frac{vcos\beta}{g} = \frac{\sqrt{2gH}}{g}$
$t = \sqrt{\frac{2H}{g}}$
Reviewed by Dr. Abubacker
Physics Educator & NEET/JEE Specialist
Ph.D. in Physics with 19+ years of teaching experience.