Question

The potential energy of a projectile at its maximum height is equal to its kinetic energy there. If the velocity of projection is $20_{ms^{-1}}$ its time of flight is $(g=10_{ms^{-2}} )$

• A. 2s
• B. $2\sqrt{2}s$
• C. $\frac{1}{2} s$
• D. $\frac{1}{\sqrt{2}}s$

Answer 1

Answer: (B)

$P.E = K.E$
$\frac{1}{2} mu^2 \, sin^2 \, \theta = \frac{1}{2} mu^2 \, cos^2 \, \theta$
then $\theta = 45^\circ $ and use $T= \frac{2U \, sin \, \theta}{g}$