Question

The speed of a projectile at the highest point becomes$\frac{1}{\sqrt{2}}$ 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

Answer: (A)

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

Answer 2

Velocity at the highest point is given by $u\cos\theta = \frac{u}{\sqrt{2}}$ (given) ∴ θ = 45^o

Horizontal range $R = \frac{u^{2}\sin 2\theta}{g} = \frac{u^{2}\sin(2 \times 45^{o})}{g} = \frac{u^{2}}{g}$