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The speed of a projectile at its maximum height is half of its initial speed. The angle of projection is
A
$60^\circ$
B
$15^\circ$
C
$30^\circ$
D
$45^\circ$
Detailed Solution
Let the initial speed be $v_0$ and the angle of projection be $\theta$.
At the maximum height the vertical component of velocity is zero; only the horizontal component remains: $v = v_0\cos\theta$
Given $v = \frac{v_0}{2}$: $\frac{v_0}{2} = v_0\cos\theta$
$\cos\theta = \frac{1}{2}$
$\theta = 60^\circ$
At the maximum height the vertical component of velocity is zero; only the horizontal component remains: $v = v_0\cos\theta$
Given $v = \frac{v_0}{2}$: $\frac{v_0}{2} = v_0\cos\theta$
$\cos\theta = \frac{1}{2}$
$\theta = 60^\circ$