A body is projected vertically upwards with speed u from the top of a tower. It strikes at the ground with speed 3u. What is (a) height of tower and (b) time take by the body to reach at the ground?

Explanation

To solve the problem, we need to analyze the motion of the body projected vertically upwards from the top of a tower. We will use the equations of motion to find both the height of the tower and the time taken to reach the ground.

### Given:
- Initial velocity \( u \) (upwards)
- Final velocity \( v = 3u \) (downwards)
- Acceleration due to gravity \( g \) (downwards)

### (a) Height of the Tower

1. **Identify the equation of motion**: We will use the third equation of motion:
\[ v^2 = u^2 + 2as \]
Here, \( v \) is the final velocity, \( u \) is the initial velocity, \( a \) is the acceleration, and \( s \) is the displacement (height of the tower, \( h \)).

2. **Substitute the known values**:
- Since the body is projected upwards, we take upwards as positive and downwards as negative. Therefore:
\[ v = -3u \quad (\text{downwards}) \]
\[ u = u \quad (\text{upwards}) \]
\[ a = -g \quad (\text{downwards}) \]
\[ s = -h \quad (\text{downwards}) \]

3. **Insert these values into the equation**:
\[ (-3u)^2 = u^2 + 2(-g)(-h) \]
\[ 9u^2 = u^2 + 2gh \]

4. **Rearrange the equation**:
\[ 9u^2 - u^2 = 2gh \]
\[ 8u^2 = 2gh \]

5. **Solve for \( h \)**:
\[ h = \frac{8u^2}{2g} = \frac{4u^2}{g} \]

### (b) Time Taken to Reach the Ground

1. **Identify the equation of motion for time**: We will use the first equation of motion:
\[ v = u + at \]

2. **Substitute the known values**:
\[ -3u = u - gt \]

3. **Rearrange the equation**:
\[ -3u - u = -gt \]
\[ -4u = -gt \]
\[ gt = 4u \]

4. **Solve for \( t \)**:
\[ t = \frac{4u}{g} \]

### Final Answers:
- (a) Height of the tower \( h = \frac{4u^2}{g} \)
- (b) Time taken to reach the ground \( t = \frac{4u}{g} \)