A man drops a ball downside from the roof of a tower of height `400` metres. At the same time another ball is thrown upside with a velocity `50 "meter"//sec` from the surface of the tower, then they will meet at which height from the surface of the tower.

Options

A. 100 metres
B. 320 metres
C. 80 metres
D. 240 meters

Explanation

To solve the problem of the two balls meeting at a certain height, we can break down the solution step by step.

### Step 1: Understand the scenario
We have a tower of height 400 meters. A ball is dropped from the top of the tower (initial velocity \( u_1 = 0 \) m/s), and another ball is thrown upwards from the base of the tower with an initial velocity \( u_2 = 50 \) m/s. We need to find the height \( x \) from the surface of the tower where the two balls meet.

### Step 2: Define the equations of motion
For the ball dropped from the roof:
- The distance traveled by the first ball after time \( t \) is given by:
\[ x = u_1 t + \frac{1}{2} a_1 t^2 \]
Here, \( u_1 = 0 \), \( a_1 = 10 \, \text{m/s}^2 \) (downward), so:
\[ x = 0 + \frac{1}{2} \cdot 10 \cdot t^2 = 5t^2 \]

For the ball thrown upwards:
- The distance traveled by the second ball after time \( t \) is given by:
\[ 400 - x = u_2 t + \frac{1}{2} a_2 t^2 \]
Here, \( u_2 = 50 \, \text{m/s} \), \( a_2 = -10 \, \text{m/s}^2 \) (upward), so:
\[ 400 - x = 50t - \frac{1}{2} \cdot 10 \cdot t^2 = 50t - 5t^2 \]

### Step 3: Set the equations equal
Since both balls meet at the same height \( x \), we can set the two equations equal to each other:
\[ x = 5t^2 \]
\[ 400 - x = 50t - 5t^2 \]

Substituting the first equation into the second gives:
\[ 400 - 5t^2 = 50t - 5t^2 \]

### Step 4: Simplify the equation
Rearranging the equation:
\[ 400 = 50t \]
\[ t = \frac{400}{50} = 8 \, \text{seconds} \]

### Step 5: Calculate the height \( x \)
Now that we have the time \( t \), we can find the height \( x \) where the two balls meet:
\[ x = 5t^2 = 5 \cdot (8^2) = 5 \cdot 64 = 320 \, \text{meters} \]

### Step 6: Find the height from the surface
Since the total height of the tower is 400 meters, the height from the surface of the tower is:
\[ \text{Height from surface} = 400 - x = 400 - 320 = 80 \, \text{meters} \]

### Final Answer
The two balls will meet at a height of **80 meters** from the surface of the tower.

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