Understanding the Problem
In this scenario, a baseball is launched with an initial velocity $v_0 = 20\text{ m/s}$ at an angle $\theta = 45^\circ$ relative to the horizontal. The player is initially standing $d = 50\text{ m}$ away. We need to find the speed and direction the player must run to catch the ball at its original height.
Step 1: Calculate the Horizontal Range of the Ball
The horizontal range $R$ of a projectile launched from and landing at the same height is given by the formula: $R = \frac{v_0^2 \sin(2\theta)}{g}$
Assuming $g \approx 9.8\text{ m/s}^2$: $R = \frac{20^2 \cdot \sin(90^\circ)}{9.8} = \frac{400 \cdot 1}{9.8} \approx 40.82\text{ m}$
Step 2: Determine the Time of Flight
The time of flight $T$ is given by: $T = \frac{2 v_0 \sin(\theta)}{g}$ $T = \frac{2 \cdot 20 \cdot \sin(45^\circ)}{9.8} = \frac{40 \cdot 0.707}{9.8} \approx 2.88\text{ s}$
Step 3: Solve for the Player's Motion
The ball lands at $40.82\text{ m}$ from the thrower. The player is standing at $50\text{ m}$. Since the ball lands closer than the player is standing, the player must run towards the thrower to catch it.
Distance to travel $\Delta x = |50 - 40.82| = 9.18\text{ m}$
Since the player must reach this point in time $T = 2.88\text{ s}$, the required speed $v_p$ is: $v_p = \frac{\Delta x}{T} = \frac{9.18}{2.88} \approx 3.19\text{ m/s}$
The direction is towards the thrower.