This chapter belongs to a previous NCERT/CBSE curriculum and is not part of the current 2026-27 chapter list. Removed in 2026-27.
Welcome to the NCERT Solutions for Class 9 Science - Chapter Gravitation. This page offers a step-by-step solution to the specific question from Exercise 6, Question 17:
A stone is allowed to fall from the top of a tower 100 m high and at the same time another stone is projected vertically upwards from the ground with a velocity of 25 m/s. Calculate when and where the two stones will meet.
. With detailed answers and explanations for each chapter, students can strengthen their understanding and prepare confidently for exams. Ideal for CBSE and other board students, this resource will simplify your study experience.A stone is allowed to fall from the top of a tower 100 m high and at the same time another stone is projected vertically upwards from the ground with a velocity of 25 m/s. Calculate when and where the two stones will meet.
Let the two stones meet after a time t.
(i) For the stone dropped from the tower:
Initial velocity, u = 0 m/s
Let the displacement = s
Acceleration due to gravity, g = 9.8 m s−2
From the equation of motion,
...(1)(ii) For the stone thrown upwards:
Initial velocity, u = 25 m/s
Let the displacement = s'.
Acceleration due to gravity, g = −9.8 m s−2
Equation of motion,
...(2)The combined displacement is ;
In 4 s, the falling stone has covered a distance given by equation (1) as
Therefore, the stones will meet after 4 s and the distance is 80 m from the top.
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:
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Welcome to the NCERT Solutions for Class 9 Science - Chapter . This page offers a step-by-step solution to the specific question from Excercise 6 , Question 17: A stone is allowed to fall from the top of a tower 100 m high and at the same time another stone is ....
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