First slide

Projection Under uniform Acceleration

Question

A particle is projected from ground with a velocity of $20 \sqrt{2} m / s$ making an angle of $60^{o}$ with horizontal. Then the time after which its velocity vector makes an angle of $45^{o}$ with horizontal is (Take $g = 10 m / s^{2}$ ).

Moderate

A

$(2 \sqrt{3} - 1) \sec$
B

$\sqrt{6} \sec$
C

$(\sqrt{5} - 1) \sec$
D

$(\sqrt{6} - \sqrt{2}) \sec$

Solution

Horizontal component of velocity remains constant.
$∴ V \cos 45^{o} = u \cos 60^{o} \Rightarrow V . \frac{1}{\sqrt{2}} = 20 \sqrt{2} . \frac{1}{2} \Rightarrow V = 20 m / s$
Now, $V \sin 45^{o} = u \sin 60^{o} - g t$
$\Rightarrow t = \frac{20 \sqrt{2} \times \frac{\sqrt{3}}{2} - 20 \times \frac{1}{\sqrt{2}}}{10} \sec = (\sqrt{6} - \sqrt{2}) \sec$

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