A ball is dropped from top of a tower 250 m high. At the same time, another ball is thrown upwards with a velocity of `50 ms^(-1)` from the ground. The time at which they cross each other is

Options

A. 3 second
B. 5 second
C. 10 second
D. 12 second

Answer

To solve the problem step by step, we will analyze the motion of both balls and find the time at which they cross each other.

### Step 1: Understand the scenario
We have two balls:
- Ball A is dropped from the top of a tower that is 250 m high.
- Ball B is thrown upwards from the ground with an initial velocity of 50 m/s.

### Step 2: Define the motion equations
1. For Ball A (dropped from the tower):
- Initial velocity (u) = 0 m/s (since it is dropped)
- Distance fallen after time t = \( s_A = \frac{1}{2} g t^2 \)
- Here, \( g \) (acceleration due to gravity) = 10 m/s².
- The distance fallen will be \( s_A = \frac{1}{2} \times 10 \times t^2 = 5t^2 \).
- The height above the ground when they meet = \( h = 250 - s_A = 250 - 5t^2 \) (Equation 1).

2. For Ball B (thrown upwards):
- Initial velocity (u) = 50 m/s.
- The distance covered upwards after time t = \( s_B = ut - \frac{1}{2} g t^2 \).
- Therefore, \( s_B = 50t - 5t^2 \) (Equation 2).

### Step 3: Set the distances equal
At the time they cross each other, the distance fallen by Ball A will be equal to the distance covered by Ball B:
- From Equation 1 and Equation 2, we have:
\[ 250 - 5t^2 = 50t - 5t^2 \]

### Step 4: Simplify the equation
We can cancel \( -5t^2 \) from both sides:
\[ 250 = 50t \]

### Step 5: Solve for time t
Now, we can solve for t:
\[ t = \frac{250}{50} = 5 \text{ seconds} \]

### Conclusion
The time at which the two balls cross each other is **5 seconds**.

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