Question

a body is projected with a speed u such that the maximum height is equal to horizontal range. find the horizontal range

Answer 1

Let us suppose that a body is projected with a velocity u with an angle $\theta$ from the horizontal.

In that case, Maximum Height (H) and Horizontal Range (R) are given by:

$\displaystyle \boxed{H = \frac{u^2\sin^2\theta}{2g}} \\ \\ \\ \boxed{R=\frac{u^2\sin 2\theta}{g}=\frac{2u^2\sin\theta\cos\theta}{g}}$

We are given that both are equal. So,

$\displaystyle H = R \\ \\ \\ \implies \frac{\cancel{u^2}\sin^{\cancel{2}}\theta}{2\cancel{g}}=\frac{2\cancel{u^2}\cancel{\sin\theta}\cos\theta}{\cancel{g}} \\ \\ \\ \implies \frac{\sin\theta}{2}= 2\cos \theta \\ \\ \\ \implies \boxed{\tan \theta = 4}$

Now, to find Range, we must find $\sin 2\theta$

We have:

$\displaystyle\sin 2\theta = \frac{2\tan \theta}{1+\tan^2\theta} \\ \\ \\ \implies \sin 2\theta = \frac{2(4)}{1+(4)^2} \\ \\ \\ \implies \boxed{\sin 2\theta = \frac{8}{17}}$

Thus, Range becomes:

$\displaystyle R = \frac{u^2\sin 2\theta}{g}\\ \\ \\ \implies R = \frac{u^2}{g} \times \frac{8}{17} \\ \\ \\ \implies \boxed{\bold{R = \frac{8u^2}{17g}}}$