Question

The horizontal range of a projectile fired at an angle of ${15}^{\circ }\text{is}50m$. If it is fired with the same speed at an angle of ${45}^{\circ }$, its range will be

• A. $100m$
• B. $60m$
• C. $141m$
• D. $71m$

Answer 1

The correct option is A $100m$

Find the horizontal Range for $\theta ={15}^{\circ }$

Given:

Angle of projection, $\theta ={15}^{\circ }$

Horizontal range,$R=50m$

Horizontal range $\left(R\right)=\frac{u^2\sin 2\theta }{g}$

By putting the values in the above relation, we get

$R=50m=\frac{u^2\sin (2\times {15}^{\circ })}{g}$

$50\times g=u^2\sin {30}^{\circ }=u^2\times \frac{1}{2}$

$50\times g\times 2=u^2$

$u^2=50\times 9.8\times 2=100\times 9.8=980m/s$

$u=\sqrt{980}=\sqrt{49\times 20}$

$u=7\times 2\times \sqrt{5}m/s$


Find the horizontal Range for $\theta ={45}^{\circ }$

For $\theta ={45}^{\circ }$

$R=\frac{u^2\sin \sin 2\times {45}^{\circ }}{g}=\frac{u^2}{g}(\because \sin \sin {90}^{\circ }=1)R=\frac{14{\sqrt{5}}^2}{g}$

$R=\frac{14\times 14\times 5}{9.8}=100m$

Final Answer: (c) $100m$