A body is thrown horizontally from the top of a tower and strikes the ground after three seconds at an angle of `45^@` with the horizontal. Find the height of the tower and the speed with which the body was projected. `(Take g = 9.8 m//s^2)`

Explanation

To solve the problem step by step, we will use the principles of projectile motion.

### Step 1: Understand the problem
A body is thrown horizontally from the top of a tower and strikes the ground after 3 seconds at an angle of 45 degrees with the horizontal. We need to find the height of the tower and the speed with which the body was projected.

### Step 2: Identify the known values
- Time of flight, \( t = 3 \) seconds
- Acceleration due to gravity, \( g = 9.8 \, \text{m/s}^2 \)
- Angle of impact with the horizontal, \( \theta = 45^\circ \)

### Step 3: Calculate the height of the tower
Using the formula for the height of a free-falling object:
\[ h = \frac{1}{2} g t^2 \]
Substituting the known values:
\[ h = \frac{1}{2} \times 9.8 \times (3)^2 \]
\[ h = \frac{1}{2} \times 9.8 \times 9 \]
\[ h = \frac{1}{2} \times 88.2 \]
\[ h = 44.1 \, \text{meters} \]

### Step 4: Calculate the final vertical velocity
The final vertical velocity \( v_y \) can be calculated using the formula:
\[ v_y = u_y + g t \]
Since the body is thrown horizontally, the initial vertical velocity \( u_y = 0 \):
\[ v_y = 0 + 9.8 \times 3 \]
\[ v_y = 29.4 \, \text{m/s} \]

### Step 5: Relate the horizontal and vertical velocities
At the moment of impact, the angle of projection is given as \( 45^\circ \). This means:
\[ \tan(45^\circ) = \frac{v_y}{v_x} = 1 \]
Thus, we have:
\[ v_y = v_x \]
From the previous step, we found \( v_y = 29.4 \, \text{m/s} \), so:
\[ v_x = 29.4 \, \text{m/s} \]

### Step 6: Conclusion
The height of the tower is \( 44.1 \, \text{meters} \) and the speed with which the body was projected horizontally is \( 29.4 \, \text{m/s} \).

### Summary of Results
- Height of the tower, \( h = 44.1 \, \text{meters} \)
- Speed of projection, \( u = 29.4 \, \text{m/s} \)