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If u is the initial velocity of a projectile and v is the velocity at any instant, then the maximum horizontal range $R_{m}$ is equal to
Explanation
Horizontal range = $u^{2}\sin 2\theta/g$ For maximum range $\theta = 45^{\circ}$ $\therefore R_{max} = u^{2}\sin 90^{\circ}/g = u^{2}/g ( \because \sin 90^{\circ} = 1)$