Question

Prove that for two angles of projections $45+\theta$ and $45-\theta$ with same velocity ${}^{\prime }{v}^{\prime }$ range is same.

Answer 1

We know that the Range $=\frac{u^2\sin 2\varphi }{g}$

For $\varphi =45+\theta$

Range $=\frac{u^2\sin 2(45+\theta )}{g}$

$=\frac{u^2\sin (90+2\theta )}{g}$

$=\frac{u^2\cos 2\theta }{g}$

Now, For $\varphi =45-\theta$

Range $=\frac{u^2\sin 2(45-\theta )}{g}$

$=\frac{u^2\sin (90-2\theta )}{g}$

$=\frac{u^2\cos 2\theta }{g}$

Hence, Proved.