A ball is thrown vertically upwards with a speed of 49 m s . Calculate maximum height of the ball.

Given Initial speed u = 49 m s Final speed at maximum height v = 0 We know that, v 2 - u 2 = 2 g h , putting the values, 0 2 - ( 49 ) 2 = 2 ( - 9 . 8 ) h ⇒ h = 122 . 5 m

A ball is thrown vertically upwards with a speed of 49 m s . Calculate total time taken by the ball to reach the earth's surface.

Given Initial velocity u = 49 m s Since the ball reaches back to us, displacement h = 0 Using, h = u t - 1 2 g t 2 0 = 49 t - 1 2 ( 9 . 8 ) t 2 ⇒ t = 0 or 10 s Since 0 time is meaningless here, so time taken to reach back is 10 s . Note: We have ignored the air friction in this calculation.

If a body is thrown vertically upwards at a speed of 10 m s then after how much time and with what velocity will it reach back to us?

Given Initial velocity v = 10 m s Since the body reaches back to the same point, its displacement is zero. Using h = u t + 1 2 g t 2 0 = 10 t - 1 2 ( 10 ) t 2 t = 0 , 2 s Since zero second is meaningless here, so time to reach back is two seconds. Now using v 2 - u 2 = 2 g h v 2 - 10 2 = 2 ( 10 ) ( 0 ) ⇒ v = 10 m s or - 10 m s The body will come back with same speed as it was thrown.

E M B I B E

Science (Theory & Practical) Class-9>Chapter 11 - Gravitation>Exercise

Chhattisgarh Board Solutions for Chapter: Gravitation, Exercise 3: Exercise

Author:Chhattisgarh Board

Chhattisgarh Board Science Solutions for Exercise - Chhattisgarh Board Solutions for Chapter: Gravitation, Exercise 3: Exercise

Attempt the practice questions on Chapter 11: Gravitation, Exercise 3: Exercise with hints and solutions to strengthen your understanding. Science (Theory & Practical) Class-9 solutions are prepared by Experienced Embibe Experts.