SELECT * FROM question_mgmt as q WHERE id=4181 AND status=1 SELECT id,question_no,question,chapter FROM question_mgmt as q WHERE courseId=8 AND subId=16 AND chapterId=190 and ex_no='6' AND status=1 ORDER BY CAST(question_no AS UNSIGNED) CBSE Free NCERT Solution of 9th science Gravitation a stone is thrown vertically upward with an initia

Question:

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?

Answer:

 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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