Question

The speed of a projectile at the maximum height is half of its initial speed $V$. Its horizontal range is

• A. $\frac{V^2}{\sqrt{3g}}$
• B. $\frac{2V^2}{\sqrt{3g}}$
• C. $\frac{\sqrt{3}}{2}\frac{V^2}{g}$
• D. $\frac{\sqrt{3}{V}^2}{g}$

Answer 1

The correct option is C $\frac{\sqrt{3}}{2}\frac{V^2}{g}$

As we know that horizontal component remains constant in projectile motion :

$V_x=\frac{V}{2}=V\cos \theta \Rightarrow [\theta ={60}^0]$
Range $=\frac{V^2\sin 2\theta }{g}=\frac{V^2\sin 120}{g}=\frac{\sqrt{3}}{2}\frac{V^2}{g}$