Question

Two stones are thrown up simultaneously from the edge of a cliff $240\, m$ high with initial speed of $10\, m/s$ and $40\, m/s$ respectively. Which of the following graph best represents the time variation of relative position of the second stone with respect to the first? (Assume stones do not rebound after hitting the ground and neglect air resistance, take $g=10\, m/s^2$)

• A.

  

• B.

  

• C.

  

• D.

  

Answer 1

The Correct Option is C

To solve this problem, we need to analyze the relative motion of the two stones thrown upwards from a cliff. Let's break it down step-by-step:

1. Initial conditions:

   • Height of cliff, h = 240\, \text{m}
   • Initial velocity of first stone, u_1 = 10\, \text{m/s}
   • Initial velocity of second stone, u_2 = 40\, \text{m/s}
   • Acceleration due to gravity, g = 10\, \text{m/s}^2

2. Equation of motion:

   For an object thrown upwards, the position as a function of time t is given by:

   s = ut - \frac{1}{2}gt^2

3. Calculate position for each stone:

   • Position of first stone, S_1: S_1 = u_1 t - \frac{1}{2} gt^2 = 10t - 5t^2
   • Position of second stone, S_2: S_2 = u_2 t - \frac{1}{2} gt^2 = 40t - 5t^2

4. Relative position:

   The relative position of the second stone with respect to the first stone is:

   \Delta S = S_2 - S_1 = (40t - 5t^2) - (10t - 5t^2) = 30t

5. Interpretation of relative position equation:

   The equation \Delta S = 30t indicates that the relative position is a linear function of time t, which means the graph of relative position vs. time would be a straight line with a positive slope.

   This matches with the graph having a linear, increasing slope starting from the origin. Hence, the correct graph should be a straight line through the origin with a positive slope.

Therefore, the graph that best represents the time variation of the relative position of the second stone with respect to the first is:

This matches the given correct answer in the options.