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 `(v_0)/(g) sqrt(v_0^2 - 4 gh))`.

Explanation

The trajectory of projectile is given by
`y = x tan theta - (gx^2)/(2 v_0^2) (1 + tan^2 theta)`
Gun is adjusted for minimum range, therefore, `prop = 45^@`
`y = x - g(x^2)/(v_0^2)`
For `y = h`, we have `h = x - (g)/(v_0^2) x^2`
`x^2 - (v_0^2)/(g) x + (v_0^2)/(g) h = 0`
If `x_1 and x_2` are roots of the above equation,
`x_1 + x_2 = (v_0^2)/(g) rArr x_1 x_2 = (v_0^2)/(g) h`
`(x_1 + x_2)^2 = (x_1 + x_2)^2 - 4 x_1 x_2 = ((v_0^2)/(g)) - 4 (v_0^2)/(g) h`
`x_1 - x_2 = (v_0)/(g) sqrt(v_0^2 - 4 gh)`.