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 of when two balls cross each other, we can use the equations of motion. Let's break down the solution step by step.

### Step 1: Understand the Scenario
We have two balls:
- Ball 1 is dropped from a height of 250 m (initial velocity \( u_1 = 0 \)).
- Ball 2 is thrown upwards from the ground with an initial velocity of \( u_2 = 50 \, \text{m/s} \).

### Step 2: Set Up the Equations of Motion
For Ball 1 (dropped from the tower):
- The displacement \( S_1 \) after time \( t \) is given by:
\[ S_1 = u_1 t + \frac{1}{2} a t^2 \]
Here, \( u_1 = 0 \) and \( a = g = 10 \, \text{m/s}^2 \) (acceleration due to gravity).
Thus, the equation simplifies to:
\[ S_1 = \frac{1}{2} g t^2 = 5 t^2 \]
- The height from the top of the tower to the position where they meet is \( 250 - S_1 \):
\[ h_1 = 250 - S_1 = 250 - 5t^2 \]

For Ball 2 (thrown upwards):
- The displacement \( S_2 \) after time \( t \) is given by:
\[ S_2 = u_2 t + \frac{1}{2} (-g) t^2 \]
Here, \( u_2 = 50 \, \text{m/s} \) and \( a = -g = -10 \, \text{m/s}^2 \).
Thus, the equation becomes:
\[ S_2 = 50t - 5t^2 \]

### Step 3: Set the Displacements Equal
At the time they cross each other, the distance covered by Ball 1 (falling down) plus the distance covered by Ball 2 (going up) should equal the height of the tower:
\[ S_1 + S_2 = 250 \]
Substituting the expressions for \( S_1 \) and \( S_2 \):
\[ 5t^2 + (50t - 5t^2) = 250 \]
This simplifies to:
\[ 50t - 5t^2 = 250 \]
Rearranging gives:
\[ 5t^2 - 50t + 250 = 0 \]

### Step 4: Solve the Quadratic Equation
Dividing the entire equation by 5:
\[ t^2 - 10t + 50 = 0 \]
Using the quadratic formula \( t = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \):
- Here, \( a = 1, b = -10, c = 50 \).
- The discriminant \( b^2 - 4ac = (-10)^2 - 4 \cdot 1 \cdot 50 = 100 - 200 = -100 \).

Since the discriminant is negative, we made a mistake in the equation setup. Let's re-evaluate.

### Step 5: Correct the Equation Setup
From the previous steps, we should have:
\[ 5t^2 + 50t - 250 = 0 \]
Dividing by 5:
\[ t^2 + 10t - 50 = 0 \]
Using the quadratic formula again:
\[ t = \frac{-10 \pm \sqrt{10^2 - 4 \cdot 1 \cdot (-50)}}{2 \cdot 1} \]
Calculating the discriminant:
\[ t = \frac{-10 \pm \sqrt{100 + 200}}{2} \]
\[ t = \frac{-10 \pm \sqrt{300}}{2} \]
\[ t = \frac{-10 \pm 10\sqrt{3}}{2} \]
\[ t = -5 \pm 5\sqrt{3} \]
Since time cannot be negative, we take:
\[ t = -5 + 5\sqrt{3} \]
Calculating \( \sqrt{3} \approx 1.732 \):
\[ t \approx -5 + 8.66 \approx 3.66 \, \text{s} \]

### Final Answer
The time at which the two balls cross each other is approximately \( 3.66 \, \text{s} \).