A body thrown horizontally from the top of a tower touches the ground at a distance of 160 m from the foot of the tower. If the velocity of projection is 40 m/s then the height of the towers is

Options

A. 98 m
B. 9.6 m
C. 58.8 m
D. 78.4 m

Answer

To solve the problem step by step, we will break it down into manageable parts:

### Step 1: Understand the Problem
A body is thrown horizontally from the top of a tower with an initial horizontal velocity of 40 m/s. It lands 160 m away from the base of the tower. We need to find the height of the tower.

### Step 2: Identify the Horizontal Motion
Since the body is thrown horizontally, we can use the formula for horizontal distance:
\[ d = u_h \cdot t \]
Where:
- \( d \) = horizontal distance (160 m)
- \( u_h \) = initial horizontal velocity (40 m/s)
- \( t \) = time of flight (in seconds)

### Step 3: Calculate the Time of Flight
Rearranging the formula to find \( t \):
\[ t = \frac{d}{u_h} \]
Substituting the known values:
\[ t = \frac{160 \, \text{m}}{40 \, \text{m/s}} = 4 \, \text{s} \]

### Step 4: Identify the Vertical Motion
Now, we need to find the height of the tower using the vertical motion. The vertical distance (height of the tower) can be calculated using the kinematic equation:
\[ h = ut + \frac{1}{2} a t^2 \]
Where:
- \( h \) = height of the tower
- \( u \) = initial vertical velocity (0 m/s, since it is thrown horizontally)
- \( a \) = acceleration due to gravity (approximately \( 10 \, \text{m/s}^2 \))
- \( t \) = time of flight (4 s)

### Step 5: Substitute Values into the Equation
Since the initial vertical velocity \( u = 0 \):
\[ h = 0 \cdot t + \frac{1}{2} g t^2 \]
Substituting the values:
\[ h = \frac{1}{2} \cdot 10 \, \text{m/s}^2 \cdot (4 \, \text{s})^2 \]
\[ h = \frac{1}{2} \cdot 10 \cdot 16 \]
\[ h = 80 \, \text{m} \]

### Final Answer
The height of the tower is **80 meters**.

---