Two balls are dropped from different heights at different instants. Seconds ball is dropped 2 sec. after the first ball. If both balls reach the ground simultaneousl after 5 sec. of dropping the first ball. then the difference of initial heights of the two balls will be :- `(g=9.8m//s^(2))`

Options

A. 58.8m
B. 78.4m
C. 98.0m
D. 117.6m

Explanation

To solve the problem, we need to find the difference in initial heights of two balls dropped from different heights at different times. Let's break it down step by step.

### Step 1: Define the variables
- Let \( g = 9.8 \, \text{m/s}^2 \) (acceleration due to gravity).
- Let \( t_1 = 5 \, \text{s} \) (time for the first ball).
- Let \( t_2 = 3 \, \text{s} \) (time for the second ball, since it is dropped 2 seconds after the first).

### Step 2: Calculate the distance fallen by the first ball
The distance fallen by the first ball, \( S_1 \), can be calculated using the formula for distance under uniform acceleration:
\[ S_1 = \frac{1}{2} g t_1^2 \]
Substituting the values:
\[ S_1 = \frac{1}{2} \times 9.8 \times (5)^2 = \frac{1}{2} \times 9.8 \times 25 \]
\[ S_1 = 4.9 \times 25 = 122.5 \, \text{m} \]

### Step 3: Calculate the distance fallen by the second ball
The distance fallen by the second ball, \( S_2 \), is given by:
\[ S_2 = \frac{1}{2} g t_2^2 \]
Substituting the values:
\[ S_2 = \frac{1}{2} \times 9.8 \times (3)^2 = \frac{1}{2} \times 9.8 \times 9 \]
\[ S_2 = 4.9 \times 9 = 44.1 \, \text{m} \]

### Step 4: Calculate the difference in heights
The difference in initial heights \( H \) between the two balls is:
\[ H = S_1 - S_2 \]
Substituting the values we calculated:
\[ H = 122.5 - 44.1 = 78.4 \, \text{m} \]

### Conclusion
The difference in initial heights of the two balls is \( 78.4 \, \text{m} \).