A cannon: If you elevate a certain cannon so that it makes an angle of $t$ degrees with the ground, then the cannonball will strike the ground $\frac{m^2 \sin (2 t)}{g}$ feet downrange, where $g=32$ feet per second is acceleration due to gravity, and $m$ is the muzzle velocity in feet per second. If the muzzle velocity is 300 feet per second, what angle would you use to make the cannonball land 1000 feet downrange?