Question

A truck is moving with velocity u. An object is projected from the truck with velocity u making an angle of 60^o with horizontal as seen from the truck. What is the angle of projection as seen from ground?

Moderate

Solution

If $α^{1}$ be the angle of projection as seen from the ground, then
$\tan α^{1} = \frac{u \sin α}{u + u \cos α} = \frac{\sin α}{1 + \cos α} = \frac{\sin 60^{o}}{1 + \cos 60^{o}} = \frac{1}{\sqrt{3}}$
$\Rightarrow α^{1} = 30^{o}$

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