First slide

Projection Under uniform Acceleration

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)

Moderate

A

2 s
B

2√2 s
C

$\frac {1}{2} s$
D

$\frac {1}{\sqrt 2}s$

Solution

KE = PE
$\large \frac{1}{2}m{u^2}{\cos ^2}\theta = \frac{1}{2}m{u^2}{\sin ^2}\theta$
$\large Tan\theta = 1,\,\theta = {45^0}$
$\large T = \frac{{2u\sin \theta }}{g} = \frac{{2 \times 20 \times \frac{1}{{\sqrt 2 }}}}{{10}} = 2\sqrt 2 \sec$

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