Question

A projectile is thrown with some initial velocity at an angle $α$ to the horizontal. Its velocity when it is at the highest point is (2/5)^1/2 times the velocity when it is at height half of the maximum height. Find the angle of projection a with the horizontal.

Moderate

Solution

$(u \cos α) = \sqrt{\frac{2}{5}} \sqrt{(u \cos α)^{2} + \{(u \sin α)^{2} - 2 g h\}}$ ….(i)
Here, $h = \frac{H}{2} = \frac{u^{2} \sin^{2} α}{4 g}$ (ii)
Solving Eqs. (i) and (ii), we get
$α = 60^{0} .$

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