Two balls are projected simultaneously with the same speed from the top of a tower-one upwards and the other downwards.If they reach the ground in 6s and 2s ,the height of the tower is

Options

A. 120 m
B. 60 m
C. 80 m
D. 30 m

Answer

To solve the problem, we need to analyze the motion of the two balls projected from the top of a tower. Let's denote the height of the tower as \( h \), the initial speed of both balls as \( u \), and the acceleration due to gravity as \( g \) (approximately \( 9.8 \, \text{m/s}^2 \)).

### Step 1: Write the equations of motion for both balls.

1. **For the ball projected upwards (Ball 1)**:
The time taken to reach the ground is \( t_1 = 6 \, \text{s} \).
Using the equation of motion:
\[ h = ut_1 - \frac{1}{2} g t_1^2 \]
Substituting \( t_1 = 6 \):
\[ h = 6u - \frac{1}{2} g (6^2) = 6u - 18g \]

2. **For the ball projected downwards (Ball 2)**:
The time taken to reach the ground is \( t_2 = 2 \, \text{s} \).
Using the equation of motion:
\[ h = ut_2 + \frac{1}{2} g t_2^2 \]
Substituting \( t_2 = 2 \):
\[ h = 2u + \frac{1}{2} g (2^2) = 2u + 2g \]

### Step 2: Set the two equations for height equal to each other.

Since both expressions represent the height of the tower \( h \), we can set them equal:
\[ 6u - 18g = 2u + 2g \]

### Step 3: Solve for \( u \).

Rearranging the equation:
\[ 6u - 2u = 18g + 2g \]
\[ 4u = 20g \]
\[ u = \frac{20g}{4} = 5g \]

### Step 4: Substitute \( u \) back into one of the height equations to find \( h \).

Using the equation for Ball 2:
\[ h = 2u + 2g \]
Substituting \( u = 5g \):
\[ h = 2(5g) + 2g = 10g + 2g = 12g \]

### Step 5: Calculate the numerical value of \( h \).

Using \( g \approx 9.8 \, \text{m/s}^2 \):
\[ h = 12 \times 9.8 \approx 117.6 \, \text{m} \]

Thus, the height of the tower is approximately \( 117.6 \, \text{m} \).

### Final Answer:
The height of the tower is \( h \approx 117.6 \, \text{m} \).

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