Question

The speed of a projectile at its maximum height is $\frac{\sqrt{3}}{2}$ times of its initial speed. If range of projectile is $p$ times the maximum height reached by it, then find the value of $p$.

• A. $4$
• B. $4\sqrt{3}$
• C. $5$
• D. $2\sqrt{3}$

Answer 1

The correct option is B $4\sqrt{3}$

Let $u$ be the initial speed of a projectile and $\theta$ be the angle of projection.
Speed at maximum height of a projectile $=u\cos \theta$
According to question
$u\cos \theta =\frac{\sqrt{3}}{2}u\Rightarrow \cos \theta =\frac{\sqrt{3}}{2}\Rightarrow \tan \theta =\frac{1}{\sqrt{3}}R=pH$ (given)
By using the relation, $R=\frac{4H}{\tan \theta }$ we get
$pH=\frac{4H}{\tan \theta }\Rightarrow p=\frac{4}{\tan \theta }=4\sqrt{3}$