Chapter 6.2, Problem 81AYU

Artillery A projectile fired into the first quadrant from the origin of a coordinate system will pass through the point $\left(x,y\right)$ at time $t$ according to the relationship $\cot \theta =\frac{2x}{2y+g{t}^2}$ where $\theta =$ the angle of elevation of the launcher and $g=$ the acceleration due to gravity $=32.2feet/secon{d}^2$. An artilleryman is firing at an enemy bunker located 2450 feet up the side of a hill that is 6175 feet away. He fires a round, and exactly $2.27$ seconds later he scores a direct hit.

(a) What angle of elevation did he use?

(b) If the angle of elevation is also given by $\sec \theta =\frac{v_{0}t}{x}$, where $v_0$ is the muzzle velocity of the weapon, find the muzzle velocity of the artillery piece he used.