To solve this problem, we'll use the equations of projectile motion to find the height at which the player at first base catches the ball. Here are the key steps:

Given:

• Distance between the two players (horizontal distance): $d = 90 \, \text{ft}$
• Initial speed of the ball: $v_0 = 50 \, \text{mph}$
• Launch angle: $\theta = 19^\circ$
• Initial height: $h_0 = 5 \, \text{ft}$
• Gravitational acceleration: $g = 32 \, \text{ft/s}^2$

First, we need to convert the initial speed from miles per hour (mph) to feet per second (ft/s):

$v_0 = 50 \, \text{mph} \times \frac{5280 \, \text{ft}}{1 \, \text{mile}} \times \frac{1 \, \text{hr}}{3600 \, \text{sec}} = 73.333 \, \text{ft/s}$

Step 1: Horizontal motion equation

The horizontal motion is governed by:

$x(t) = v_0 \cos(\theta) t$

We can solve for the time, $t$, when the ball reaches first base (i.e., $x = 90 \, \text{ft}$):

$90 = 73.333 \cos(19^\circ) t$

Step 2: Vertical motion equation

The vertical motion is governed by:

$y(t) = h_0 + v_0 \sin(\theta) t - \frac{1}{2} g t^2$

Now, we'll solve for the time of flight and substitute it into the vertical equation to find the height at which the ball is caught. I'll calculate it step-by-step.The player at first base catches the ball at a height of approximately $9.033 \, \text{ft}$ above the ground.

Would you like any further details or explanations?

Here are five related questions you could explore:

1. How would the answer change if the initial speed were different?
2. What if the angle of projection was steeper, say 30°? How would that affect the outcome?
3. How long does it take for the ball to reach first base?
4. What is the maximum height the ball reaches during its flight?
5. How does air resistance impact projectile motion in real-world scenarios?

Tip: In projectile motion problems, always break the motion into horizontal and vertical components to simplify calculations.