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. 20 m
B. 80 m
C. 170 m
D. 40 m

Explanation

To solve the problem step by step, we will use the kinematic equation for the distance fallen under gravity:

\[ h = \frac{1}{2} g t^2 \]

where:
- \( h \) is the height from which the ball is dropped,
- \( g \) is the acceleration due to gravity (approximately \( 9.8 \, \text{m/s}^2 \)),
- \( t \) is the time in seconds.

### Step 1: Identify the time taken by each ball
- The first ball is dropped and takes \( t_1 = 5 \) seconds to reach the ground.
- The second ball is dropped 2 seconds later, which means it falls for \( t_2 = 5 - 2 = 3 \) seconds.

### Step 2: Calculate the height from which the first ball is dropped
Using the formula for height:
\[ h_1 = \frac{1}{2} g t_1^2 \]
Substituting \( g = 9.8 \, \text{m/s}^2 \) and \( t_1 = 5 \) seconds:
\[ h_1 = \frac{1}{2} \times 9.8 \times (5)^2 \]
\[ h_1 = \frac{1}{2} \times 9.8 \times 25 \]
\[ h_1 = 4.9 \times 25 \]
\[ h_1 = 122.5 \, \text{meters} \]

### Step 3: Calculate the height from which the second ball is dropped
Using the same formula:
\[ h_2 = \frac{1}{2} g t_2^2 \]
Substituting \( g = 9.8 \, \text{m/s}^2 \) and \( t_2 = 3 \) seconds:
\[ h_2 = \frac{1}{2} \times 9.8 \times (3)^2 \]
\[ h_2 = \frac{1}{2} \times 9.8 \times 9 \]
\[ h_2 = 4.9 \times 9 \]
\[ h_2 = 44.1 \, \text{meters} \]

### Step 4: Calculate the difference in heights
Now, we find the difference in heights:
\[ \Delta h = h_1 - h_2 \]
\[ \Delta h = 122.5 - 44.1 \]
\[ \Delta h = 78.4 \, \text{meters} \]

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

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