A hose pipe lying on the ground shoots a stream of water upward at an angle `60^@` to the horizontal at a speed of `20 ms^(-1)`. The water strikes a wall 20m away at a height of `(g=10ms^2)`

Options

A. 14.64 m
B. 7.32 m
C. 29.28 m
D. 10 m

Answer

To solve the problem, we need to find the vertical height at which the water strikes the wall, given the initial speed, angle, and horizontal distance.

### Step-by-Step Solution:

1. **Identify the Given Values:**
- Initial speed \( u = 20 \, \text{m/s} \)
- Angle of projection \( \theta = 60^\circ \)
- Horizontal distance to the wall \( x = 20 \, \text{m} \)
- Acceleration due to gravity \( g = 10 \, \text{m/s}^2 \)

2. **Use the Projectile Motion Equation:**
The vertical height \( y \) at a horizontal distance \( x \) can be calculated using the projectile motion formula:
\[ y = x \tan \theta - \frac{g x^2}{2 u^2 \cos^2 \theta} \]

3. **Calculate \( \tan \theta \) and \( \cos \theta \):**
- \( \tan 60^\circ = \sqrt{3} \)
- \( \cos 60^\circ = \frac{1}{2} \)

4. **Substitute Values into the Equation:**
Substitute \( x = 20 \, \text{m} \), \( g = 10 \, \text{m/s}^2 \), \( u = 20 \, \text{m/s} \), \( \tan 60^\circ = \sqrt{3} \), and \( \cos 60^\circ = \frac{1}{2} \):
\[ y = 20 \tan 60^\circ - \frac{10 \cdot 20^2}{2 \cdot 20^2 \cdot \left(\frac{1}{2}\right)^2} \]
\[ y = 20 \sqrt{3} - \frac{10 \cdot 400}{2 \cdot 400 \cdot \frac{1}{4}} \]

5. **Simplify the Equation:**
- The second term simplifies as follows:
\[ \frac{10 \cdot 400}{2 \cdot 400 \cdot \frac{1}{4}} = \frac{10 \cdot 400}{200} = 20 \]
- Therefore, the equation becomes:
\[ y = 20 \sqrt{3} - 20 \]

6. **Calculate \( y \):**
- Now, we can calculate \( y \):
\[ y = 20 (\sqrt{3} - 1) \]
- Using \( \sqrt{3} \approx 1.732 \):
\[ y \approx 20 (1.732 - 1) = 20 \times 0.732 \approx 14.64 \, \text{m} \]

### Final Answer:
The vertical height at which the water strikes the wall is approximately \( 14.64 \, \text{m} \).