Use the model for projectile motion, assuming there is no air resistance and *g* = 32 feet per second per second.

A baseball, hit 3 feet above the ground, leaves the bat at an angle of 45° and is caught by an outfielder 3 feet above the ground and 300 feet from home plate. What is the initial speed of the ball, and how high does it rise? (Round your answers to two decimal places.)

initial speed

ft/s

maximum height

We can solve this projectile motion problem by breaking it into its horizontal and vertical components.

### Step 1: Horizontal motion equation

The horizontal distance \( x \) traveled by the baseball is given by:

\[
x = v_0 \cos(\theta) t
\]

where:
- \( x = 300 \, \text{feet} \) (horizontal distance),
- \( v_0 \) is the initial speed,
- \( \theta = 45^\circ \) is the launch angle,
- \( t \) is the total time of flight.

Since \( \cos(45^\circ) = \sin(45^\circ) = \frac{\sqrt{2}}{2} \), the equation becomes:

\[
300 = v_0 \frac{\sqrt{2}}{2} t
\]

### Step 2: Vertical motion equation

The vertical displacement equation is:

\[
y = v_0 \sin(\theta) t - \frac{1}{2} g t^2
\]

where:
- \( y = 0 \) (since the ball is hit and caught at 3 feet above the ground),
- \( g = 32 \, \text{ft/s}^2 \) (acceleration due to gravity),
- \( v_0 \sin(\theta) = v_0 \frac{\sqrt{2}}{2} \) (vertical component of initial velocity).

Substituting, we get:

\[
0 = v_0 \frac{\sqrt{2}}{2} t - \frac{1}{2} (32) t^2
\]

Simplifying:

\[
0 = v_0 \frac{\sqrt{2}}{2} t - 16 t^2
\]

Factoring out \( t \):

\[
t \left( v_0 \frac{\sqrt{2}}{2} - 16 t \right) = 0
\]

Thus, either \( t = 0 \) (the moment the ball is hit) or:

\[
v_0 \frac{\sqrt{2}}{2} = 16 t
\]

Solving for \( t \):

\[
t = \frac{v_0 \frac{\sqrt{2}}{2}}{16}
\]

### Step 3: Solve for \( v_0 \)

We now have two expressions for \( t \):

1. From the horizontal motion:
\[
t = \frac{300}{v_0 \frac{\sqrt{2}}{2}}
\]

2. From the vertical motion:
\[
t = \frac{v_0 \frac{\sqrt{2}}{2}}{16}
\]

Equating these:

\[
\frac{300}{v_0 \frac{\sqrt{2}}{2}} = \frac{v_0 \frac{\sqrt{2}}{2}}{16}
\]

Simplifying:

\[
300 \times \frac{v_0 \frac{\sqrt{2}}{2}}{1} = \frac{(v_0 \frac{\sqrt{2}}{2})^2}{16}
\]

Multiplying through by 16:

\[
4800 = \left(v_0 \frac{\sqrt{2}}{2}\right)^2
\]

Taking the square root:

\[
\sqrt{4800} = v_0 \frac{\sqrt{2}}{2}
\]

Solving for \( v_0 \):

\[
v_0 = \frac{2 \sqrt{4800}}{\sqrt{2}} = 2 \sqrt{2400}
\]

Finally:

\[
v_0 \approx 97.98 \, \text{ft/s}
\]

### Step 4: Maximum height

The maximum height occurs when the vertical velocity is zero. Using the vertical component of the velocity \( v_y = v_0 \sin(\theta) \), we can find the time to reach the maximum height:

\[
v_y = v_0 \frac{\sqrt{2}}{2}
\]

The time to reach maximum height is:

\[
t_{\text{max}} = \frac{v_0 \frac{\sqrt{2}}{2}}{g} = \frac{97.98 \times \frac{\sqrt{2}}{2}}{32} \approx 2.16 \, \text{seconds}
\]

The maximum height is given by:

\[
h_{\text{max}} = y_0 + v_0 \frac{\sqrt{2}}{2} t_{\text{max}} - \frac{1}{2} g t_{\text{max}}^2
\]

Substituting the values:

\[
h_{\text{max}} = 3 + (97.98 \times \frac{\sqrt{2}}{2}) \times 2.16 - \frac{1}{2} \times 32 \times (2.16)^2
\]

Solving for \( h_{\text{max}} \):

\[
h_{\text{max}} \approx 3 + 149.52 - 74.79 = 77.73 \, \text{feet}
\]

### Final answers:

- **Initial speed**: \( 97.98 \, \text{ft/s} \)
- **Maximum height**: \( 77.73 \, \text{feet} \)

Would you like to see more details or have any questions about these steps?

---

1. What is the equation for horizontal motion in projectile motion?
2. How does the angle of projection affect both vertical and horizontal velocity components?
3. Why is gravity considered a negative acceleration in vertical motion?
4. How do you derive the time of flight in projectile motion?
5. What factors influence the maximum height reached in a projectile motion?

**Tip**: Always decompose the initial velocity into horizontal and vertical components when solving projectile motion problems.

A baseball, hit 3 feet above the ground, leaves the bat at an angle of 45° and is caught by an outfielder 3 feet above the ground and 300 feet from home plate. What is the initial speed of the ball, and how high does it rise?

Solve a real-world projectile motion problem involving a baseball hit at 45° and caught 300 feet away. Learn how to calculate the initial speed (97.98 ft/s) and the maximum height (77.73 feet) of the ball using kinematic equations and trigonometry. Suitable for high school students.

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