A baseball thrown at an angle of 65.0 ∘
 above the horizontal strikes a building 18.0 m
 away at a point 5.00 m
 above the point from which it is thrown. Ignore air resistance. Find the magnitude of the velocity of the baseball just before it strikes the building.

To solve this problem, we need to use the principles of projectile motion. The steps involved are as follows:

### Step 1: Break down the problem
Given:
- Angle of projection, \(\theta = 65.0^\circ\)
- Horizontal distance (range), \(R = 18.0 \, \text{m}\)
- Vertical displacement, \(\Delta y = 5.00 \, \text{m}\)

We need to find the magnitude of the velocity of the baseball just before it strikes the building.

### Step 2: Horizontal and vertical components of initial velocity
The initial velocity can be broken into horizontal (\(v_{0x}\)) and vertical (\(v_{0y}\)) components:
\[
v_{0x} = v_0 \cos\theta
\]
\[
v_{0y} = v_0 \sin\theta
\]
where \(v_0\) is the initial velocity of the baseball.

### Step 3: Time of flight
The time of flight can be determined using the horizontal motion equation:
\[
R = v_{0x} \times t
\]
Substituting \(v_{0x}\):
\[
t = \frac{R}{v_0 \cos\theta}
\]

### Step 4: Vertical displacement equation
Using the vertical motion equation:
\[
\Delta y = v_{0y} \times t - \frac{1}{2} g t^2
\]
where \(g\) is the acceleration due to gravity (\(9.8 \, \text{m/s}^2\)).

Substituting \(t\) and the expressions for \(v_{0x}\) and \(v_{0y}\):
\[
\Delta y = v_0 \sin\theta \times \frac{R}{v_0 \cos\theta} - \frac{1}{2} g \left(\frac{R}{v_0 \cos\theta}\right)^2
\]
Simplifying this:
\[
\Delta y = R \tan\theta - \frac{g R^2}{2 v_0^2 \cos^2\theta}
\]
Solving this equation for \(v_0\):
\[
v_0^2 = \frac{g R^2}{2 \cos^2\theta (R \tan\theta - \Delta y)}
\]

### Step 5: Final velocity before impact
The magnitude of the final velocity just before impact can be found using:
\[
v_f = \sqrt{v_{fx}^2 + v_{fy}^2}
\]
where
- \(v_{fx} = v_{0x} = v_0 \cos\theta\) (since horizontal velocity remains constant)
- \(v_{fy} = v_{0y} - gt\)

First, solve for \(v_0\), then calculate \(v_{fx}\) and \(v_{fy}\), and finally, find \(v_f\).

Let's calculate these step by step.

### Calculation

1. **Calculate \(v_0\):**

\[
v_0^2 = \frac{(9.8)(18.0)^2}{2 \cos^2(65.0^\circ) \left(18.0 \tan(65.0^\circ) - 5.00\right)}
\]

2. **Calculate \(v_{fx}\) and \(v_{fy}\):**

\[
v_{fx} = v_0 \cos(65.0^\circ)
\]

\[
v_{fy} = v_0 \sin(65.0^\circ) - g \times t
\]

3. **Calculate \(v_f\):**

\[
v_f = \sqrt{v_{fx}^2 + v_{fy}^2}
\]

Now, I'll compute the numerical values.The magnitude of the velocity of the baseball just before it strikes the building is approximately **12.91 m/s**.

Would you like to go into further detail or have any questions about the steps? Here are 5 related questions to consider:

1. How would the result change if the angle of projection was different?
2. What if the initial speed of the baseball was known—how could we find the time of flight?
3. How does air resistance impact the velocity in real-world scenarios?
4. What would be the time of flight for this problem?
5. How would the maximum height reached by the baseball be calculated?

**Tip:** When dealing with projectile motion, always start by breaking down the motion into horizontal and vertical components, as it simplifies the calculations.

A baseball thrown at an angle of 65.0° above the horizontal strikes a building 18.0 m away at a point 5.00 m above the point from which it is thrown. Ignore air resistance. Find the magnitude of the velocity of the baseball just before it strikes the building.

Find out how to calculate the final velocity of a baseball thrown at an angle of 65.0° above the horizontal. This problem involves key concepts of projectile motion, including kinematics and trigonometry. Suitable for students in Grades 11-12.

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