A ball projected upwards from the foot of a tower. The ball crosses the top of the tower twice after an interval of `6 s` and the ball reaches the ground after `12 s`. The height of the tower is `(g = 10 m//s^2)` :

Options

A. 120 m
B. 135 m
C. 175 m
D. 180 m

Explanation

To solve the problem, we can break it down into several steps:

### Step 1: Understand the Motion of the Ball
The ball is projected upwards and crosses the top of the tower twice. The time taken to cross the top of the tower twice is 6 seconds, and the total time of flight until it reaches the ground is 12 seconds.

### Step 2: Determine Initial Velocity (U)
Using the total time of flight, we can apply the equation of motion:
\[ S = Ut + \frac{1}{2} a t^2 \]
Since the ball returns to the ground, the displacement \( S = 0 \), the time \( t = 12 \) seconds, and the acceleration \( a = -g = -10 \, \text{m/s}^2 \).

Substituting these values into the equation:
\[ 0 = U(12) + \frac{1}{2}(-10)(12^2) \]
\[ 0 = 12U - 5(144) \]
\[ 12U = 720 \]
\[ U = \frac{720}{12} = 60 \, \text{m/s} \]

### Step 3: Determine Velocity at the Top of the Tower (V)
Now, we can find the velocity of the ball when it crosses the top of the tower. The time taken to cross the top of the tower is 6 seconds. We can use the same equation of motion:
\[ S = Ut + \frac{1}{2} a t^2 \]
Here, \( t = 6 \) seconds, and we need to find the height of the tower \( H \).

Using the initial velocity \( U = 60 \, \text{m/s} \):
\[ H = 60(6) + \frac{1}{2}(-10)(6^2) \]
\[ H = 360 - 5(36) \]
\[ H = 360 - 180 = 180 \, \text{m} \]

### Step 4: Use the Velocity at the Top of the Tower
Now, we can find the velocity \( V \) when the ball crosses the top of the tower:
Using the equation:
\[ V = U + at \]
\[ V = 60 + (-10)(6) \]
\[ V = 60 - 60 = 0 \, \text{m/s} \]

### Step 5: Calculate the Height of the Tower
Now, we can use the velocities \( U \) and \( V \) to find the height \( H \) of the tower using the equation:
\[ V^2 - U^2 = 2aH \]
Substituting \( V = 0 \) and \( U = 30 \, \text{m/s} \):
\[ 0^2 - 30^2 = 2(-10)H \]
\[ -900 = -20H \]
\[ H = \frac{900}{20} = 45 \, \text{m} \]

### Final Answer
The height of the tower is \( 45 \, \text{m} \).