Answer to Question #148058 in Physics for buddy

Question

A quarterback throws a ball at 16.6 m/s at an angle of 32.1 degrees above the
horizontal. His wide receiver beside him leaves at the same time the ball is thrown. How
fast must he move to catch the ball?

Answer 1

Let's write the equations of motion of the ball in horizontal and vertical directions:

$$x=v_0tcos\theta, (1)$$ $$y=v_0tsin\theta-\dfrac{1}{2}gt^2, (2)$$

here, $x$ is the horizontal displacement of the ball (or the range), $v_0=16.6\ \dfrac{m}{s}$ is the initial velocity of the ball, $t$ is the time of flight of the ball, $\theta=32.1^{\circ}$ is the launch angle, $y$ is the vertical displacement of the ball (or the height) and $g=9.8\ \dfrac{m}{s^2}$ is the acceleration due to gravity.

Let's first find the time of flight of the ball from the second equation:

$$0=8.82t-4.9t^2,$$ $$4.9t^2-8.82t=0,$$ $$t=\dfrac{8.82}{4.9}=1.8\ s.$$

Then, substituting $t$ into the first equation we can find the range of the ball:

$$x=16.6\ \dfrac{m}{s}\cdot 1.8\ s\cdot cos32.1^{\circ}=25.3\ m.$$

Unfortunately, from the condition of the question we don't know the distance between the quarterback and receiver. Let's assume that when the quarterback throws the ball, the distance between him and the receiver is 16 meters. Then, in order to catch the ball, the receiver must run away from the quarterback with the speed:

$$v=\dfrac{x}{t}=\dfrac{25.3\ m-16\ m}{1.8\ s}=5.2\ \dfrac{m}{s}.$$

Answer:

$v=5.2\ \dfrac{m}{s}.$