Answer:
The equations for the y position of the trajectory is:
[tex]y=sin\phi v_0t-\frac{1}{2}gt^2[/tex]
The range is of the trajectory is reached at y = 0:
[tex]sin\phi v_0t-\frac{1}{2}gt^2 = 0[/tex]
Solving for time t:
[tex]t=\frac{2sin\phi v_0}{g}[/tex]
The x position of the trajectory is given by:
[tex]x=cos\phi v_0t[/tex]
Combining the equations we get the function:
[tex]x=\frac{2sin\phi cos\phi v_0^2}{g}=sin(2\phi)\frac{v_0^2}{g}[/tex]
Taking the derivative and setting it to zero:
[tex]\frac{dx}{d\phi}=2cos(2\phi)\frac{v_0^2}{g}= 0[/tex]
Find the maximum angle:
[tex]\phi=45[/tex]
Using this solution to find the trajectory in terms of x and y and setting it equal to height h:
[tex]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[/tex]
Normalize:
[tex]x^2-\frac{v_0^2}{g}x+\frac{hv_0^2}{g}=0[/tex]
Use quadratic formula:
[tex]x_{1/2}=\frac{v_0^2}{2g}+/-v_0\sqrt{\frac{v_0^2-4hg}{4g^2}}[/tex]
The distance is the difference between the two points:
[tex]d=|x_1-x_2|[/tex]
[tex]d=\frac{v_0}{g} \sqrt{v_0^2-4hg}[/tex]
