Presentation on theme: "Physics. Session Kinematics - 3 Session Objectives Problems ? Free fall under gravity."— Presentation transcript:
Session Kinematics - 3
Session Objectives Problems ? Free fall under gravity
Free Fall Under Gravity - g g is constant = - g -y Motion under gravity is along y-axis. Downward is negative
One dimensional equations of motion under gravity v = u - gt v² = u² - 2gs Distance traveled in n th second of fall
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)
Solution Hence answer is (c)
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
Solution If u is the velocity of balloon in the upward direction when the stone was dropped, then –76 = 6u – 180
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.
Solution a(2) = –6 × = –10 ms –2 (ii) = –75 m (i)
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 )
Solution 2t 2 – 20t + 25 = 0
Class Exercise - 5 For an object moving in one dimension, find a if v = 3x 2 + 4x m/s. Solution :
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 ?
Solution v cos = u m
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.
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
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.)
Solution v 1 : v 2 = 1 : 2 Given that t 2 = t 1
Class Exercise - 9 Along X-axis: Along Y-axis: Solution :
Class Exercise - 10 Find x and y if t = 5 s. Use the same method as question 9. Solution :