SELECT * FROM question_mgmt as q WHERE id=4180 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 released from the top of a tower of hei

Question:

A stone is released from the top of a tower of height 19.6 m. Calculate its final velocity.

Answer:

 Initial velocity of the stone = u = 0

 Final velocity of the stone = v

 Height of the stone = s = 19.6 m

 Acceleration due to gravity = g = 9.8 m s−2

According to the equation of motion under gravity:

v2 − u2 = 2 gs

v2 − 02 = 2 × 9.8 × 19.6

v2 = 2 × 9.8 × 19.6 = (19.6)2

v = 19.6 m s−1

Hence, the velocity of the stone just before touching the ground is 19.6 m s−1.


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