On flat terrain, a projectile with an initial velocity of $190$ (ft/sec) is fired at a $20$ - foot high vertical wall $1000$ feet from the tip of the cannon inclined at an angle a with the horizontal.
We know that:
$x(t, s, a) = s \,t \cos a$
$y(t, s, a) = -16 t^2 + s\, t \sin a$
Steps I took:
Find when the projectile hits the ground:
$y(t, s, a) = 0$
$t = \frac{s \sin a}{16}$
For a given muzzle speed $s$ and angle $a$ , the horizontal range of the cannon is:
$f(a) = x(\frac{1}{16}\cdot 190 \sin a, 190, a)= \frac{9025 (\cos a)( \sin a)}{4}$
When $f'(a) = 0$,
$a = \frac{3\pi}{4} \text{ or } \frac{\pi}{4}$
So projectile hits or soars over the wall at angles $\frac{3\pi}{4} \text{ or } \frac{\pi}{4}$. However, this answer is wrong and I am clueless as to how to go about finding the correct answer. What steps do I need to take to find the correct answer?