Are ads bothering you?

NCERT Questions

NCERT Question 1

NCERT Question 2 Important

NCERT Question 3

NCERT Question 4

NCERT Question 5 Important

NCERT Question 6 Important

NCERT Question 7

NCERT Question 8 Important

NCERT Question 9

NCERT Question 10

NCERT Question 11 Important

NCERT Question 12 Important

NCERT Question 13 Important

NCERT Question 14 Important

NCERT Question 15 You are here

NCERT Question 16

NCERT Question 17 Important

NCERT Question 18 Important

NCERT Question 19 Important

NCERT Question 20 Important

NCERT Question 21 Important

NCERT Question 22

Class 9

Chapter 10 Class 9 - Gravitation (Term 2)

Last updated at May 30, 2019 by Teachoo

NCERT Question 15 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 & total distance covered by the stone? Given Initial velocity = u = 49 m/s At the highest point of motion, velocity becomes 0 Therefore Final velocity = v = 0 m/s And Acceleration = −g = −10 m/s2 (Acceleration is negative because stone is thrown upwards) We need to find the maximum height (s) We know u, v & a, finding s using 3rd equation of motion v2 − u2 = 2as 02 − 402 = 2 × (−10) × s –1600= –20 × s 1600 = 20 × s 20 × s = 1600 s = 1600/20 s = 80 m Therefore, maximum height attained by the ball would be 80 m Now, when we throw stone upwards It reaches highest point and comes down Now, when we throw stone upwards It reaches highest point and comes down So, initial and final point is same ∴ Net Displacement = 0 m And Total Distance covered = 2 × Maximum Height covered by stone = 2 × 80 = 160 m