Question

A projectile is fired with velocity $v_0$ from a gun adjusted for a maximum range. It passes through two points $P$ and $Q$ whose heights above the horizontal are $h$ each. Show that the separation of the two points is $\frac{v_0}{g}\sqrt{v_0^2-4gh}$

Answer 1

The trajectory of projectile is given by

$y=x\tan \theta -\frac{g{x}^2}{2v_0^2}(1+{\tan }^2\theta )$

Gun is adjusted for minimum range ,therefore ,$\alpha ={45}^o$

$y=x-g\frac{x^2}{v_0^2}$

For $y=h$, we have $h=x-\frac{8}{v_0^2}{x}^2$

$x^2-\frac{v_0^2}{g}x+\frac{v{[}_0^2}{g}h=0$

If $x_1$ and $x^2$ $=\frac{v_0^2}{g}\Rightarrow x_1{x}_2=\frac{v_0^2}{g}h$

$(x_1+x_2{)}^2=(x_1+x_2{)}^2-4x_1{x}_2=\left(\frac{v_0^2}{g}\right)-4\frac{v_0^2}{g}h$

$x_1-x_2=\frac{v_0}{g}\sqrt{v_0^2-4gh}$