The highest barrier that a projectile can clear is 13.5 m, when the projectile is launched at an angle of `15.0^(@)` above the horizontal. What is the projectile's launch speed?

Options

A. 15.7 m/s
B. 62.8 m/s
C. 44.4 m/s
D. 22.2 m/s

Answer

To find the launch speed of the projectile that can clear a maximum height of 13.5 meters when launched at an angle of 15 degrees above the horizontal, we can use the following steps:

### Step-by-Step Solution:

1. **Identify the Known Values**:
- Maximum height (H) = 13.5 m
- Launch angle (θ) = 15 degrees
- Acceleration due to gravity (g) = 9.8 m/s²

2. **Use the Formula for Maximum Height**:
The formula for the maximum height (H) reached by a projectile is given by:
\[ H = \frac{u^2 \sin^2 \theta}{2g} \]
where \( u \) is the initial launch speed, \( \theta \) is the launch angle, and \( g \) is the acceleration due to gravity.

3. **Rearranging the Formula**:
We can rearrange the formula to solve for \( u^2 \):
\[ u^2 = \frac{2gH}{\sin^2 \theta} \]

4. **Substituting the Known Values**:
Substitute \( H = 13.5 \) m, \( g = 9.8 \) m/s², and \( \theta = 15 \) degrees into the equation. First, calculate \( \sin(15^\circ) \):
\[ \sin(15^\circ) \approx 0.2588 \]
Now, calculate \( \sin^2(15^\circ) \):
\[ \sin^2(15^\circ) \approx (0.2588)^2 \approx 0.0675 \]
Now substitute into the rearranged formula:
\[ u^2 = \frac{2 \times 9.8 \times 13.5}{0.0675} \]

5. **Calculating \( u^2 \)**:
First calculate the numerator:
\[ 2 \times 9.8 \times 13.5 = 264.6 \]
Now divide by \( \sin^2(15^\circ) \):
\[ u^2 = \frac{264.6}{0.0675} \approx 3925.93 \]

6. **Finding \( u \)**:
Take the square root to find \( u \):
\[ u = \sqrt{3925.93} \approx 62.7 \text{ m/s} \]

7. **Final Answer**:
The launch speed of the projectile is approximately \( 62.7 \) m/s.

### Summary of the Steps:
1. Identify known values.
2. Use the maximum height formula.
3. Rearrange to solve for initial speed.
4. Substitute known values into the equation.
5. Calculate \( u^2 \).
6. Take the square root to find \( u \).
7. State the final answer.