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.