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)`

Options

A. `h= 44.1 m`, `u_x = 9.4 m//s`
B. `h= 41 m`, `u_x = 29.4 m//s`
C. `h= 44.1 m`, `u_x = 29.4 m//s`
D. `h= 44.1 m`, `u_x = 9.4 m//s`

Explanation

To solve the problem step by step, we will analyze the motion of the body thrown horizontally from the top of the tower.

### Step 1: Understand the motion
The body is thrown horizontally, meaning its initial vertical velocity (Uy) is 0. The body falls under the influence of gravity, which accelerates it downwards with an acceleration of \( g = 9.8 \, \text{m/s}^2 \).

### Step 2: Calculate the height of the tower
The time taken to hit the ground is given as \( t = 3 \, \text{s} \). We can use the formula for vertical displacement (height of the tower, \( h \)):
\[ h = U_y t + \frac{1}{2} g t^2 \]
Since the initial vertical velocity \( U_y = 0 \):
\[ h = 0 \cdot t + \frac{1}{2} g t^2 = \frac{1}{2} \cdot 9.8 \cdot (3^2) \]
Calculating this:
\[ h = \frac{1}{2} \cdot 9.8 \cdot 9 = \frac{1}{2} \cdot 88.2 = 44.1 \, \text{m} \]

### Step 3: Find the vertical velocity just before impact
The vertical velocity \( V_y \) just before hitting the ground can be calculated using:
\[ V_y = U_y + g t \]
Again, since \( U_y = 0 \):
\[ V_y = 0 + 9.8 \cdot 3 = 29.4 \, \text{m/s} \]

### Step 4: Determine the horizontal velocity
The problem states that the angle of impact with the horizontal is \( 45^\circ \). This means that the horizontal and vertical components of the velocity are equal:
\[ V_x = V_y \]
Thus,
\[ V_x = 29.4 \, \text{m/s} \]

### Step 5: Find the initial speed with which the body was projected
Since the body was thrown horizontally, the initial horizontal speed \( U_x \) is equal to \( V_x \):
\[ U_x = V_x = 29.4 \, \text{m/s} \]

### Final Results
- Height of the tower: \( h = 44.1 \, \text{m} \)
- Speed with which the body was projected: \( U = 29.4 \, \text{m/s} \)