First slide

Projection Under uniform Acceleration

Question

A particle is projected from ground making an angle of $\tan^{- 1} \sqrt{2}$ with horizontal. Find the angle between its velocity vector and horizontal ground when its height above the ground will be equal to half of the maximum height attained by it.

Easy

A

$60^{o}$
B

$30^{o}$
C

$45^{o}$
D

$37^{o}$

Solution

Maximum height, $H = \frac{u^{2} \sin^{2} α}{2 g}$
where $α$ is the angle of projection.
At height $\frac{H}{2}, V \cos θ = u \cos α$
and ${(V \sin θ)}^{2} = {(u \sin α)}^{2} - 2. g . \frac{H}{2}$
$\Rightarrow {(V \sin θ)}^{2} = u^{2} \sin^{2} α - g (\frac{u^{2} \sin^{2} α}{2 g}) = \frac{u^{2} \sin^{2} α}{2}$
$\Rightarrow V \sin θ = u \sin α / \sqrt{2}$
$∴ \tan θ = \frac{V \sin θ}{V \cos θ} = \frac{\tan α}{\sqrt{2}} = \frac{\sqrt{2}}{\sqrt{2}} = 1 \Rightarrow α = 45^{o}$

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