A man is standing on top of a building 100 m high. He throws two ball vertically, one at t=0 and after a time interval (less than 2 seconds). The later ball is thrown at a velocity of half the first. At t=2, both the balls reach to their and second ball is +15m.
Q. The speed of first ball is

Options

A. 20 m/s
B. 10 m/s
C. 5 m/s
D. 15 m/s

Explanation

To solve the problem step by step, we will analyze the motion of both balls and use the given information to find the speed of the first ball.

### Step 1: Define Variables
Let:
- \( V_1 \) = initial speed of the first ball
- \( V_2 = \frac{1}{2} V_1 \) = initial speed of the second ball
- \( g = 10 \, \text{m/s}^2 \) (acceleration due to gravity)
- \( h_1 \) = maximum height reached by the first ball
- \( h_2 \) = maximum height reached by the second ball

### Step 2: Maximum Height Formula
The maximum height reached by an object thrown vertically can be calculated using the formula:
\[ h = \frac{u^2}{2g} \]
where \( u \) is the initial velocity and \( g \) is the acceleration due to gravity.

### Step 3: Calculate Maximum Heights
For the first ball:
\[ h_1 = \frac{V_1^2}{2g} \]
For the second ball:
\[ h_2 = \frac{V_2^2}{2g} = \frac{\left(\frac{1}{2} V_1\right)^2}{2g} = \frac{V_1^2}{8g} \]

### Step 4: Use the Given Condition
At \( t = 2 \) seconds, the vertical gap between the first and second ball is given as 15 m:
\[ h_1 - h_2 = 15 \]

Substituting the expressions for \( h_1 \) and \( h_2 \):
\[ \frac{V_1^2}{2g} - \frac{V_1^2}{8g} = 15 \]

### Step 5: Simplify the Equation
To simplify the left side, find a common denominator:
\[ \frac{4V_1^2}{8g} - \frac{V_1^2}{8g} = 15 \]
This simplifies to:
\[ \frac{3V_1^2}{8g} = 15 \]

### Step 6: Solve for \( V_1^2 \)
Multiply both sides by \( 8g \):
\[ 3V_1^2 = 15 \times 8g \]
Substituting \( g = 10 \):
\[ 3V_1^2 = 15 \times 80 \]
\[ 3V_1^2 = 1200 \]
\[ V_1^2 = \frac{1200}{3} = 400 \]

### Step 7: Calculate \( V_1 \)
Taking the square root:
\[ V_1 = \sqrt{400} = 20 \, \text{m/s} \]

### Final Answer
The speed of the first ball is:
\[ \boxed{20 \, \text{m/s}} \]