Question

The speed of a projectile at the highest point becomes $1/\sqrt{2}$ times its initial speed. The horizontal range of the projectile will be

• A. $\frac{u^2}{g}$
• B. $\frac{u^2}{2g}$
• C. $\frac{u^2}{3g}$
• D. $\frac{u^2}{4g}$

Answer 1

The correct option is A $\frac{u^2}{g}$

Lets consider, $u=$ initial speed of the projectile

The speed of projectile at the highest point becomes $1/\sqrt{2}$ times its initial speed.

$ucos\theta =\frac{u}{\sqrt{2}}$

$cos\theta =\frac{1}{\sqrt{2}}$

$\theta ={45}^0$

Horizontal range in the projectile motion,

$R=\frac{u^{2}sin2\theta }{g}$

$R=\frac{u^{2}sin{90}^0}{g}$

$R=\frac{u^2}{g}$

The correct option is A.