Download presentation

Presentation is loading. Please wait.

Published byKristofer Doswell Modified about 1 year ago

1
Physics

2
Session Kinematics - 3

3
Session Objectives Problems ? Free fall under gravity

4
Free Fall Under Gravity - g g is constant = - g -y Motion under gravity is along y-axis. Downward is negative

5
One dimensional equations of motion under gravity v = u - gt v² = u² - 2gs Distance traveled in n th second of fall

6
Class Exercise

7
Class Exercise -1 Which of the following graphs defines the motion of an object thrown vertically up with a velocity V from a height H above the ground? (a) (b) (c) (d)

8
Solution Hence answer is (c)

9
Class Exercise -2 A stone is dropped from a rising balloon at a height of 76 m above the ground and reaches the ground in 6 s. What is the approximate velocity of the balloon when the stone was dropped? (c) 3 m/s (d) 9.8 m/s

10
Solution If u is the velocity of balloon in the upward direction when the stone was dropped, then –76 = 6u – 180

11
Class Exercise - 3 A particle moves in a straight line with a velocity v = –3t 2 + 2t + 5 ms –1, where t is the time. Assuming x(t) = 0 at t = 0, find (i) acceleration a(t), at t = 2 s, (ii) position of particle at t = 5 s.

12
Solution a(2) = –6 × = –10 ms –2 (ii) = –75 m (i)

13
Class Exercise - 4 What is the time taken by an object thrown vertically upwards at 50 m/s to attain a displacement which is half of the maximum displacement? (Take g = 10 m/s 2 )

14
Solution 2t 2 – 20t + 25 = 0

15
Class Exercise - 5 For an object moving in one dimension, find a if v = 3x 2 + 4x m/s. Solution :

16
Class Exercise - 6 m A mass m is being pulled up using two strings passing over two pulleys so that the downward velocities of the strings are equal to u. (see figure). What is the speed v of the mass ?

17
Solution v cos = u m

18
Class Exercise - 7 Two objects are thrown, one upwards and one downwards with a speed u from the top of a tower. If the time of flights are t 1 and t 2, find the time of flight of an object dropped from the tower.

19
Solution The time of flight to start from P and come back to that point is t 1 – t 2 for the ball thrown upwards. Hence, time of flight for a ball that is dropped

20
Class Exercise - 8 An object is thrown upwards from the top of a tower with a velocity v 1. At the same instance another object is thrown from the bottom of the tower with a velocity v 2. If they meet at the top of the tower, find the ratio v 1 : v 2. (Assume v 2 is just enough to reach the top of the tower.)

21
Solution v 1 : v 2 = 1 : 2 Given that t 2 = t 1

22
Class Exercise - 9 Along X-axis: Along Y-axis: Solution :

23
Class Exercise - 10 Find x and y if t = 5 s. Use the same method as question 9. Solution :

24
Thank you

Similar presentations

© 2017 SlidePlayer.com Inc.

All rights reserved.

Ads by Google