Two balls are projected upward simultaneously with speeds 40 m/s and 60 m/s. Relative position (x) of second ball w.r.t. first ball at time t = 5 s is [Neglect air resistance].

Options

A. 20 m
B. 80 m
C. 100 m
D. 120 m

Answer

To solve the problem of finding the relative position of the second ball with respect to the first ball at time \( t = 5 \) seconds, we will follow these steps:

### Step 1: Understand the initial conditions
- The first ball is projected upward with an initial speed \( u_1 = 40 \, \text{m/s} \).
- The second ball is projected upward with an initial speed \( u_2 = 60 \, \text{m/s} \).
- We need to find the relative position of the second ball with respect to the first ball after \( t = 5 \) seconds.

### Step 2: Calculate the distance traveled by the first ball
Using the equation of motion:
\[ s_1 = u_1 t - \frac{1}{2} g t^2 \]
where \( g \) (acceleration due to gravity) is approximately \( 10 \, \text{m/s}^2 \).

Substituting the values:
\[ s_1 = 40 \times 5 - \frac{1}{2} \times 10 \times (5)^2 \]
\[ s_1 = 200 - \frac{1}{2} \times 10 \times 25 \]
\[ s_1 = 200 - 125 \]
\[ s_1 = 75 \, \text{m} \]

### Step 3: Calculate the distance traveled by the second ball
Using the same equation of motion:
\[ s_2 = u_2 t - \frac{1}{2} g t^2 \]

Substituting the values:
\[ s_2 = 60 \times 5 - \frac{1}{2} \times 10 \times (5)^2 \]
\[ s_2 = 300 - \frac{1}{2} \times 10 \times 25 \]
\[ s_2 = 300 - 125 \]
\[ s_2 = 175 \, \text{m} \]

### Step 4: Calculate the relative position of the second ball with respect to the first ball
The relative position \( x \) of the second ball with respect to the first ball is given by:
\[ x = s_2 - s_1 \]
Substituting the distances we calculated:
\[ x = 175 - 75 \]
\[ x = 100 \, \text{m} \]

### Final Answer
The relative position of the second ball with respect to the first ball at \( t = 5 \) seconds is \( 100 \, \text{m} \).

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