The speed of a projectile at its maximum height is half of its initial speed. The angle of projection is
The speed of a projectile at its maximum height is half of its initial speed. The angle of projection is
- $60^{\circ}$
- $15^{\circ}$
- $30^{\circ}$
- $45^{\circ}$
Solution
The speed of a projectile at its maximum height
$\begin{aligned}
\mathrm{v}^{\prime} & =\mathrm{v}_0 \cos \theta \\
\frac{\mathrm{v}_0}{2} & =\mathrm{v}_0 \cos \theta \\
\cos \theta & =\frac{1}{2} \\
\theta & =60^{\circ}
\end{aligned}$
Asked in: NEET 2010 (Mains)
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