Question

The horizontal range of a projectile is maximum when the angle of projection is

• A. $0^o$
• B. ${30}^o$
• C. ${45}^o$
• D. ${60}^o$

Answer 1

Answer: (C)

${45}^o$

Let Velocity is $V$ and the angle of projection with horizontal is $\theta$.

$\therefore \text{Range, R}=\frac{2V^{2}sin\theta cos\theta }{g}=\frac{V^{2}sin2\theta }{g}$

$\Rightarrow \frac{dR}{d\theta }=\frac{V^2}{g}2cos2\theta =0$ $\Rightarrow 2\theta ={90}^o$ $\Rightarrow \theta ={45}^o$

$\Rightarrow \frac{d^{2}R}{d{\theta }^2}=-\frac{V^2}{g}4sin2\theta$, at $\theta ={45}^o$ $\Rightarrow \frac{d^{2}R}{d{\theta }^2}=-\frac{4V^2}{g}<0$

Hence, at $\theta ={45}^o$ , Range is maximum.