Question

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

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

Answer 1

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

Given $\frac{\sqrt{3}}{2}u=u\cos \theta$

speed at maximum height or $\cos \theta =\frac{\sqrt{3}}{2}$ or $\theta ={30}^o$

Given that $P{H}_{max}=R$

We know $H_{max}=\frac{R\tan \theta }{4}$

$P=\frac{R}{H_{max}}=\frac{4}{\tan \theta }$

$=\frac{4}{\tan {30}^o}=4\sqrt{3}$