First slide

Projection Under uniform Acceleration

Question

Two bodies are thrown up, one at 45° and the other at 60°, to the horizontal. If both bodies attain the same vertical height, then the ratio of projected speed of the first projectile to that of the second projectile is

Easy

A

$\sqrt{\frac{2}{3}}$
B

$\frac{2}{\sqrt{3}}$
C

$\sqrt{\frac{3}{2}}$
D

$\frac{\sqrt{3}}{2}$

Solution

$\begin{array}{l} H_{1} = H_{2} \\ \frac{U_{1}^{2} \sin^{2} 45}{2 g} = \frac{U_{2}^{2} \sin^{2} 60}{2 g} \\ \frac{U_{1}}{U_{2}} = \frac{\sin 60^{\circ}}{\sin 45^{\circ}} = \frac{\sqrt{3} / 2}{1 / \sqrt{2}} = \sqrt{\frac{3}{2}} \end{array}$

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