Question

A cannon placed at the origin fires a projectile with velocity ~v0, which passes during its trajectory through two points both a distance h above the horizontal, before landing. Show that if the cannon is adjusted for maximum range and air resistance is neglected, then the separation of the two points is d = v0 g p v 2 0 − 4gh.

Answer 1

Answer:

The equations for the y position of the trajectory is:

$y=sin\phi v_0t-\frac{1}{2}gt^2$

The range is of the trajectory is reached at y = 0:

$sin\phi v_0t-\frac{1}{2}gt^2 = 0$

Solving for time t:

$t=\frac{2sin\phi v_0}{g}$

The x position of the trajectory is given by:

$x=cos\phi v_0t$

Combining the equations we get the function:

$x=\frac{2sin\phi cos\phi v_0^2}{g}=sin(2\phi)\frac{v_0^2}{g}$

Taking the derivative and setting it to zero:

$\frac{dx}{d\phi}=2cos(2\phi)\frac{v_0^2}{g}= 0$

Find the maximum angle:

$\phi=45$

Using this solution to find the trajectory in terms of x and y and setting it equal to height h:

$y=\frac{sin45 v_0}{cos45v_0}x-\frac{g}{2cos^245v_0^2}x^2=x-\frac{g}{v_0^2}x^2=h$

Normalize:

$x^2-\frac{v_0^2}{g}x+\frac{hv_0^2}{g}=0$

Use quadratic formula:

$x_{1/2}=\frac{v_0^2}{2g}+/-v_0\sqrt{\frac{v_0^2-4hg}{4g^2}}$

The distance is the difference between the two points:

$d=|x_1-x_2|$

$d=\frac{v_0}{g} \sqrt{v_0^2-4hg}$