Question

Two balls $A$ and $B$ are thrown with speeds $u$ and $\frac{u}{\sqrt{2}}$ respectively. Both the balls cover the same horizontal distance before returning to the plane of projection. If the angle of projection of balls $B$ is ${15}^0$ with the horizontal, then the angle of projection of $A$ is $\theta =x^o$, Find $x$?

Answer 1

Given initial velocites of A and B are $u$ and $u/\sqrt{2}$ respectively.

And we know horizontal range $\left(R\right)=\frac{u^2\sin 2\theta }{g}$

Here, we have

$R_1=R_2$

$\Rightarrow \frac{u^{2}sin{30}^0}{g}=\frac{(u/\sqrt{2}{)}^2\sin 2\theta }{g}$

$\Rightarrow \frac{1}{2}=\frac{\sin 2\theta }{2}$

$\Rightarrow sin2\theta =1$

$\Rightarrow \theta ={45}^0$