From a balloon rising vertically with uniform acceleration f a ball is dropped, 4 seconds after this another ball is dropped from the balloon. The distance. between the two balls 2 seconds after the second ball is dropped is

Options

A. 16 (g+f)
B. 10 (g+f)
C. 8 (g+f)
D. None

Answer

To solve the problem, we need to calculate the distance between two balls dropped from a balloon that is rising with uniform acceleration. Let's break down the solution step-by-step.

### Step 1: Define the Variables
- Let \( f \) be the uniform acceleration of the balloon.
- Let \( g \) be the acceleration due to gravity (approximately \( 9.81 \, \text{m/s}^2 \)).
- The first ball is dropped at \( t = 4 \) seconds.
- The second ball is dropped at \( t = 6 \) seconds (2 seconds after the first ball).

### Step 2: Calculate the Height of the First Ball After 6 Seconds
The first ball is in free fall after being dropped. The height of the first ball after 6 seconds (2 seconds after the second ball is dropped) can be calculated using the equation of motion:
\[ h_1 = h_0 + v_0 t + \frac{1}{2} a t^2 \]
Where:
- \( h_0 \) is the initial height of the first ball when it was dropped.
- \( v_0 \) is the initial velocity of the first ball, which is equal to the velocity of the balloon at \( t = 4 \) seconds.
- \( a \) is the acceleration due to gravity (acting downwards, hence it will be negative).

#### Finding \( v_0 \) at \( t = 4 \) seconds:
The velocity of the balloon at \( t = 4 \) seconds is given by:
\[ v_0 = f \cdot t = f \cdot 4 \]

#### Height Calculation:
Now, substituting the values into the height formula for the first ball:
\[ h_1 = 0 + (f \cdot 4) \cdot 2 + \frac{1}{2} (-g) \cdot (2)^2 \]
\[ h_1 = 8f - 2g \]

### Step 3: Calculate the Height of the Second Ball After 2 Seconds
The second ball is also in free fall after being dropped. The height of the second ball after 2 seconds can be calculated similarly:
\[ h_2 = h_0 + v_0 t + \frac{1}{2} a t^2 \]
Where:
- The initial height \( h_0 \) is the height of the balloon at \( t = 6 \) seconds.
- The initial velocity \( v_0 \) is the velocity of the balloon at \( t = 6 \) seconds.

#### Finding \( v_0 \) at \( t = 6 \) seconds:
\[ v_0 = f \cdot t = f \cdot 6 \]

#### Height Calculation for the Second Ball:
\[ h_2 = 0 + (f \cdot 6) \cdot 2 + \frac{1}{2} (-g) \cdot (2)^2 \]
\[ h_2 = 12f - 2g \]

### Step 4: Calculate the Distance Between the Two Balls
Now, we can find the distance \( d \) between the two balls:
\[ d = h_1 - h_2 \]
Substituting the heights we calculated:
\[ d = (8f - 2g) - (12f - 2g) \]
\[ d = 8f - 2g - 12f + 2g \]
\[ d = -4f \]

### Conclusion
The distance between the two balls 2 seconds after the second ball is dropped is \( 4f \) in the downward direction.