First slide

Projection Under uniform Acceleration

Question

A body is projected at an angle $α$ with horizontal. It crosses a horizontal line above the point of projection during its upward journey making an angle of $θ = 45^{o}$ with horizontal. If velocity of projection is 20 m/s, and $α = 60^{o},$ find the height of the line above the point of projection.

Moderate

A

$10 \sqrt{2} m$
B

10 m
C

$10 \sqrt{3} m$
D

20 m

Solution

$\begin{array}{l} V \cos 45^{0} = 20 \cos 60^{0} \Rightarrow V = 10 \sqrt{2} m / s . \\ ∴ {(V \sin 45^{0})}^{2} = {(u \sin α)}^{2} - 2 g h . \\ \begin{array}{l} 2 g h = u^{2} \sin^{2} α - V^{2} \sin^{2} 45 \\ = 400 \times \frac{3}{4} - 200 \times \frac{1}{2} \\ ∴ h = 10 m \end{array} \end{array}$

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