A ball is dropped from the top of a building of height 80 m. At same instant another ball is thrown upwards with speed 50 m/s from the bottom of the building. The time at which balls will meet is

Options

A. 1.6 s
B. 5 s
C. 8 s
D. 10 s

Answer

To solve the problem of when the two balls will meet, we will use the equations of motion. Let's break it down step by step.

### Step 1: Define the variables
- Let the height of the building be \( H = 80 \, \text{m} \).
- Let the initial velocity of the ball dropped from the top be \( u_1 = 0 \, \text{m/s} \) (since it is dropped).
- Let the initial velocity of the ball thrown upwards be \( u_2 = 50 \, \text{m/s} \).
- Let \( g = 9.8 \, \text{m/s}^2 \) (acceleration due to gravity).
- Let \( t \) be the time in seconds when the two balls meet.

### Step 2: Write the equations of motion for both balls
1. For the ball dropped from the top:
\[ h_1 = u_1 t + \frac{1}{2} g t^2 = 0 + \frac{1}{2} g t^2 = \frac{1}{2} g t^2 \]
Thus, the distance fallen by the first ball is:
\[ h_1 = \frac{1}{2} g t^2 \]

2. For the ball thrown upwards:
\[ h_2 = u_2 t - \frac{1}{2} g t^2 = 50t - \frac{1}{2} g t^2 \]
Thus, the distance covered by the second ball is:
\[ h_2 = 50t - \frac{1}{2} g t^2 \]

### Step 3: Set up the equation for when both balls meet
The total height of the building is equal to the sum of the distances traveled by both balls:
\[ h_1 + h_2 = H \]
Substituting the expressions for \( h_1 \) and \( h_2 \):
\[ \frac{1}{2} g t^2 + (50t - \frac{1}{2} g t^2) = 80 \]
This simplifies to:
\[ 50t = 80 \]

### Step 4: Solve for \( t \)
Now, we can solve for \( t \):
\[ t = \frac{80}{50} = 1.6 \, \text{s} \]

### Conclusion
The time at which both balls will meet is \( t = 1.6 \, \text{seconds} \).

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