Two stones are thrown up simultaneously | Class 11 Physics Chapter Motion in a straight Line, Motion in a straight Line NCERT Solutions

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Academic session 2026-27, Chapter 2 in the current curriculum.

Welcome to the NCERT Solutions for Class 11 Physics - Chapter Motion in a straight Line. This page offers a step-by-step solution to the specific question from Exercise 1, Question 26:

Two stones are thrown up simultaneously from the edge of a cliff 200 m high with initial speeds of 15 m s–1 and 30 m s–1. Verify that the graph shown in Fig. 3.27 correctly represents the time variation of the relative position of the second stone with respect to the first. Neglect air resistance and assume that the stones do not rebound after hitting the ground. Take g = 10 m s–2. Give the equations for the linear and curved parts of the plot.

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Question 26:

Two stones are thrown up simultaneously from the edge of a cliff 200 m high with initial speeds of 15 m s–1 and 30 m s–1. Verify that the graph shown in Fig. 3.27 correctly represents the time variation of the relative position of the second stone with respect to the first. Neglect air resistance and assume that the stones do not rebound after hitting the ground. Take g = 10 m s–2. Give the equations for the linear and curved parts of the plot.

Answer:

For first stone:

Initial velocity, uI = 15 m/s

Acceleration, a = –g = – 10 m/s2

Using the relation,

x1 = x0 + u1t  +  ½ at2

Where, height of the cliff , x0 = 200m

x1 =  200  +  15t  -   5t2             ...... (i)

When this stone hits the ground, x1 = 0

∴– 5t2 + 15t + 200 = 0

t2 – 3t – 40 = 0

t2 – 8t + 5t – 40 = 0

t (t – 8) + 5 (t – 8) = 0

t = 8 s or t = – 5 s

Since the stone was projected at time t = 0, the negative sign before time is meaningless.

∴t = 8 s

 

For second stone:

Initial velocity, uII = 30 m/s

Acceleration, a = –g = – 10 m/s2

Using the relation,

x2 = x0 + ullt  +  ½ at2

= 200 + 30t - 5t2         ..... (ii)

At the moment when this stone hits the ground; x2 = 0

– 5t2 + 30 t + 200 = 0

t2 – 6t – 40 = 0

t2 – 10t + 4t + 40 = 0

t (t – 10) + 4 (t – 10) = 0

t (t – 10) (t + 4) = 0 t

= 10 s or t = – 4 s

Here again, the negative sign is meaningless.

∴t = 10 s

Subtracting equations (i) and (ii), we get

x2 - x1 = (200 + 30t - 5t2)  -  ( 200 + 15t + – 5t2)

x2 - x1  =  15t              .... (iii)

Equation (iii) represents the linear path of both stones.

Due to this linear relation between (x2 – x1) and t, the path remains a straight line till 8 s.

Maximum separation between the two stones is at t = 8 s.

(x2 – x1)max  = 15× 8 = 120 m

This is in accordance with the given graph.

After 8 s, only second stone is in motion whose variation with time is given by the quadratic equation: x2 – x1 = 200 + 30t – 5t2

Hence, the equation of linear and curved path is given by

x2 – x1 = 15t (Linear path)

x2 – x1 = 200 + 30t – 5t2 (Curved path)


Study Tips for Answering NCERT Questions:

NCERT questions are designed to test your understanding of the concepts and theories discussed in the chapter. Here are some tips to help you answer NCERT questions effectively:

  • Read the question carefully and focus on the core concept being asked.
  • Reference examples and data from the chapter when answering questions about Motion in a straight Line.
  • Review previous year question papers to get an idea of how such questions may be framed in exams.
  • Practice answering questions within the time limit to improve your speed and accuracy.
  • Discuss your answers with your teachers or peers to get feedback and improve your understanding.

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Welcome to the NCERT Solutions for Class 11 Physics - Chapter . This page offers a step-by-step solution to the specific question from Excercise 1 , Question 26: Two stones are thrown up simultaneously from the edge of a cliff 200 m high with initial speeds of 1....

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