Question

show that when the horizontal range is maximum the maximum height attained by the body is 1/4th of the maximum horizontal range.
pls help...

Answer 1

We know the horizontal range of the projectile,

$R=\dfrac {u^2\sin (2\theta)}{g}$

For maximum horizontal range,

$R_{\text{max}}=\dfrac {u^2}{g}\quad\iff\quad\theta=45^{\circ}$

So, the maximum height,

$h=\dfrac {u^2\sin^2\theta}{2g}\\\\h=\dfrac {u^2\cdot\dfrac {1}{2}}{2g}\\\\h=\dfrac {u^2}{4g}\\\\h=\dfrac {1}{4}\cdot\dfrac {u^2}{g}\\\\h=\dfrac {R_{\text {max}}}{4}$

Hence Proved!