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 can follow these steps:

### Step 1: Understand the motion of the balls
Both balls are projected upward with initial speeds:
- Ball 1 (first ball): \( u_1 = 40 \, \text{m/s} \)
- Ball 2 (second ball): \( u_2 = 60 \, \text{m/s} \)

Both balls experience the same downward acceleration due to gravity \( g = 9.81 \, \text{m/s}^2 \).

### Step 2: Calculate the position of each ball at \( t = 5 \) seconds
The position of an object under uniform acceleration can be calculated using the equation:
\[ s = ut + \frac{1}{2} a t^2 \]
where:
- \( s \) is the displacement,
- \( u \) is the initial velocity,
- \( a \) is the acceleration (negative in this case, since it's downward),
- \( t \) is the time.

**For Ball 1:**
\[ s_1 = u_1 t - \frac{1}{2} g t^2 \]
Substituting the values:
\[ s_1 = 40 \times 5 - \frac{1}{2} \times 9.81 \times (5)^2 \]
\[ s_1 = 200 - \frac{1}{2} \times 9.81 \times 25 \]
\[ s_1 = 200 - 122.625 \]
\[ s_1 = 77.375 \, \text{m} \]

**For Ball 2:**
\[ s_2 = u_2 t - \frac{1}{2} g t^2 \]
Substituting the values:
\[ s_2 = 60 \times 5 - \frac{1}{2} \times 9.81 \times (5)^2 \]
\[ s_2 = 300 - \frac{1}{2} \times 9.81 \times 25 \]
\[ s_2 = 300 - 122.625 \]
\[ s_2 = 177.375 \, \text{m} \]

### Step 3: Calculate the relative position
The relative position \( x \) of the second ball with respect to the first ball is given by:
\[ x = s_2 - s_1 \]
Substituting the values:
\[ x = 177.375 - 77.375 \]
\[ x = 100 \, \text{m} \]

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

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