A stone is thrown vertically upward with an initial velocity of 40 m/s. Taking g = 10 m/s2, find the maximum height reached by the stone. What is the net displacement and the total distance covered by the stone?
Initial velocity of the stone = u = 40 m/s
Final velocity of the stone = v = 0
Height of the stone = s
Acceleration due to gravity = g = −10 m s−2
Let h = maximum height attained by the stone.
According to the equation of motion under gravity:
v2 − u2 = 2 gs
Therefore, total distance covered by the stone during its upward and downward journey = 80 + 80 = 160 m
Net displacement of the stone during its upward and downward journey
= 80 + (−80) = 0
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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 15: A stone is thrown vertically upward with an initial velocity of 40 m/s. Taking g = 10 m/s2, find the....
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