NCERT Question 14 - Chapter 10 Class 9 - Gravitation

Last updated at May 30, 2019 by Teachoo

Transcript

NCERT Question 14 A stone is released from the top of a tower of height 19.6 m. Calculate its final velocity just before touching the ground. Since the stone is released from top, Initial Velocity = u = 0 m/s Distance travelled by the stone = Height of the tower = 19.6 m And, Acceleration = a = g = 9.8 m/s2 (Acceleration is positive because stone falls downwards) Finding final velocity (v) We know u, s and a, Finding v using 3rd equation of motion v2 − u2 = 2as v2 − (0) 2 = 2 × 9.8 × 19.6 v2 = 19.6 × 19.6 v2 = 19.62 v = 19.6 m/s Final velocity of the ball is 19.6 m/s

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Chapter 10 Class 9 - Gravitation

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Davneet Singh

Davneet Singh is a graduate from Indian Institute of Technology, Kanpur. He has been teaching from the past 9 years. He provides courses for Maths and Science at Teachoo.