The speed of a projectile at the highest point becomes 1√2times its initi

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

Options:

A . u2g

B . u22g

C . u23g

D . u24g

Answer: Option A
:
A
Velocity at the highest point is given by

u c o s θ = \frac{u}{\sqrt{2}} (g i v e n) ∴ θ = 45^{\circ}

Horizontal range

R = \frac{u^{2} s i n 2 θ}{g} = \frac{u^{2} s i n (2 \times 45^{\circ})}{g} = \frac{u^{2}}{g}

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