Worked example 3.4: Hail Mary pass

Question: The Longhorns are down by 4 points with 5s left in the fourth quarter. Chris Simms launches a Hail Mary pass into the end-zone, 60 yards away, where B.J. Johnson is waiting to make the catch. Suppose that Chris throws the ball at 55 miles per hour. At what angle to the horizontal must the ball be launched in order for it to hit the receiver? Neglect the effect of air resistance.
 
Answer: The formula for the horizontal range $R$ of a projectile thrown with initial velocity $v_0$ at an angle $\theta$ to the horizontal is [see Eq. (77)]:

$$R= \frac{v_0^{ 2}}{g} \sin 2\theta.$$

In this case, $R= 60\times 3\times 0.3048 = 54.86 {\rm m}$ and $v_0 = 55\times 5280\times 0.3048/3600 = 24.59 {\rm m/s}$. Hence,

$$\theta = \frac{1}{2} \sin^{-1}\left(\frac{R g}{v_0^{ 2}}\... ...\left(\frac{54.86\times 9.81}{(24.59)^2}\right) = 31.45^\circ.$$

Thus, the ball must be launched at $31.45^\circ$ to the horizontal. (Actually, $58.56^\circ$ would work just as well. Why is this?)