A football quarterback is moving straight backward at a speed of 2.2 m/s when he throws a pass to a player 24 m straight downfield. The ball is thrown at an angle of 24° relative to the ground and is caught at the same height as it is released. What is magnitude of the initial velocity of the ball relative to the quarterback in m/s? 


To solve this problem, we need to find the magnitude of the initial velocity of the ball relative to the quarterback. Here’s how we can proceed step by step:

### Step 1: Break down the motion into components
The ball is thrown at an angle of \( 24^\circ \) relative to the ground. We know two things about the motion:
- The horizontal distance to the receiver is 24 m.
- The ball is caught at the same height as it was released, meaning we can treat it as projectile motion.

### Step 2: Use projectile motion formulas
We need to use the following equations of motion:

- Horizontal displacement:  
  \[
  x = v_{0x} \cdot t
  \]
  Where:
  - \( x \) is the horizontal distance (24 m),
  - \( v_{0x} = v_0 \cdot \cos(\theta) \) is the horizontal component of the initial velocity,
  - \( t \) is the time the ball is in the air.

- Vertical motion (since the ball lands at the same height):  
  \[
  y = v_{0y} \cdot t - \frac{1}{2} g t^2 = 0
  \]
  Where:
  - \( v_{0y} = v_0 \cdot \sin(\theta) \) is the vertical component of the initial velocity,
  - \( g = 9.8 \, \text{m/s}^2 \) is the acceleration due to gravity.

### Step 3: Solve for time in the air \( t \)
Using the vertical motion equation, we can find the time \( t \) that the ball is in the air:

\[
t = \frac{2 v_0 \sin(\theta)}{g}
\]

### Step 4: Use the horizontal motion to find \( v_0 \)
Substitute this expression for \( t \) into the horizontal displacement equation:

\[
x = v_0 \cdot \cos(\theta) \cdot \frac{2 v_0 \sin(\theta)}{g}
\]

Simplifying this equation:

\[
x = \frac{2 v_0^2 \sin(\theta) \cos(\theta)}{g}
\]

Using the trigonometric identity \( \sin(2\theta) = 2 \sin(\theta) \cos(\theta) \), we get:

\[
x = \frac{v_0^2 \sin(2\theta)}{g}
\]

Solving for \( v_0 \):

\[
v_0 = \sqrt{\frac{x g}{\sin(2\theta)}}
\]

Now, substituting the known values (\( x = 24 \, \text{m}, \, \theta = 24^\circ, \, g = 9.8 \, \text{m/s}^2 \)):

\[
v_0 = \sqrt{\frac{24 \cdot 9.8}{\sin(48^\circ)}}
\]

\[
v_0 \approx \sqrt{\frac{235.2}{0.7431}} \approx \sqrt{316.5} \approx 17.79 \, \text{m/s}
\]

### Step 5: Account for the quarterback's motion
The quarterback is moving backward at 2.2 m/s, so we need to find the velocity of the ball relative to him. The velocity of the ball relative to the quarterback is:

\[
v_{\text{relative}} = v_0 - v_{\text{quarterback}}
\]

\[
v_{\text{relative}} = 17.79 - 2.2 = 15.59 \, \text{m/s}
\]

### Final Answer:
The magnitude of the initial velocity of the ball relative to the quarterback is **15.59 m/s**.

Would you like more details or have any questions?

Here are 5 related questions:
1. How would the problem change if the ball was thrown from a different height?
2. What if the quarterback was running forward instead of backward?
3. How would the angle of the throw affect the initial velocity needed?
4. What if air resistance was considered in this problem?
5. How would the time of flight be calculated if the pass was caught at a higher point?

**Tip:** When dealing with projectile motion problems, always break the motion into horizontal and vertical components, and use trigonometric identities when possible to simplify calculations.

A football quarterback is moving straight backward at a speed of 2.2 m/s when he throws a pass to a player 24 m straight downfield. The ball is thrown at an angle of 24° relative to the ground and is caught at the same height as it is released. What is magnitude of the initial velocity of the ball relative to the quarterback in m/s?

Learn how to calculate the initial velocity of a football thrown by a quarterback moving at 2.2 m/s, using projectile motion principles. This problem involves kinematics, trigonometric identities, and velocity relative to the quarterback, and is ideal for students in Grades 11-12.

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