A wide receiver is standing $40$ ft to the left ($x=-40ft$) of the quarterback who is standing at the origin. The receiver immediately accelerates at $\displaystyle 9.79 \frac{ft}{s^2}$ at an unknown angle and catches the football directly in front of the quarterback. If the quarterback counts to $3$ (seconds) before launching the ball, what is the unknown angle? What are the receiver's coordinates at the moment he catches the ball? Air resistance is neglected.
Other info: The quarterback throws the ball at a constant velocity of $\displaystyle 80.67 \frac{ft}{s}$. Acceleration due to gravity is $\displaystyle -32 \frac{ft}{s^2}$
This is a question for a math class. I have spent around $4$ hours trying to figure it out.
Hints: Let $\theta$ be the angle between the receiver's path and the $+x$ axis. The receiver's position at time $t$ is $\frac 12\cdot 9.79 t^2(\cos \theta,\sin \theta)-(40,0)$ We have that the receiver's position at the catch has $x(t)=0$ We have to assume the altitude of launch are altitude of reception are the same, though this is not specified. Let $\phi$ be the elevation angle of the release of the ball. Measure $z$ from the elevation of the release and catch. The ball position is $(0, 80.67 \cos \phi t, 80.67 \sin \phi t - \frac 12 \cdot 32 t^2)$ This gives three equations in three unknowns. Not an easy one.