First slide

Projectile motion

Question

The speed of a projectile at its highest point is v₁ and at the point half the maximum height is v₂. If $\frac{v_{1}}{v_{2}} = \sqrt{\frac{2}{5}},$ then find the angle of projection

Moderate

A

45°
B

30°
C

37°
D

60°

Solution

$\begin{array}{l} v_{1} = v_{2} \cos β = ucos θ \\ 0 = {(v_{2} \sin β)}^{2} - 2 g (H / 2) \Rightarrow v_{1}^{2} = v_{2}^{2} - gH \end{array}$
Also $\frac{v_{1}}{v_{2}} = \cos β = \sqrt{\frac{2}{5}}$

From above $v_{1} = \sqrt{\frac{2 gH}{3}}, v_{2} = \sqrt{\frac{5 gH}{3}}$
$H = \frac{u^{2} \sin^{2} θ}{2 g} \Rightarrow \sin^{2} θ = \frac{2 {gHcos}^{2} θ}{v_{1}^{2}}$
or $\tan^{2} θ = \frac{2 gH}{v_{1}^{2}} or \tan^{2} θ = \frac{2 gH \times 3}{2 gH}$
or $\tan θ = \sqrt{3} or θ = 60^{\circ}$

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