1. Physics

2. Session Kinematics - 5

3. Session Opener How far can this cannon fire the shot? Courtesy : http://physics.nad.ru/Physics/English/thermo.htm

4. Session Objectives

5. 1.Projectile motion (Definition) 2.Parameters of Projectile Motion 3.Path of Projectile Motion 4.Instantaneous radius of Curvature 5.Projectile motion on an inclined plane

6. Projectile motion (Definition) Projectile Motion = Uniform horizontal motion + Non-Uniform vertical motion (constant acceleration due to gravity) Assumptions : (1) g=constant (y<<R earth ) (2) v<<v escape (3) Air friction & force of buoyancy are neglected (4) Object is a particle

7. Projectile Motion Courtesy : www.physicsclassroom.com Gravity free path: Vertical free fall: V x =constant V y = f(t)

8. Projectile motion g=constant Y X v tx =vcos v tY = vsin  - gt H R  v

9. Parameters of Projectile Motion R x y H

10. Path of Projectile Motion

11. Class Exercise

12. Class Exercise - 1 A stone is thrown with a velocity of 19.6 m/s at an angle 30° above horizontal from the top of a building 14.7 m high. Find (i) the time after which the stone strikes the ground (ii) the distance of the landing point of the stone from the building (iii) the velocity with which the stone hits the ground (iv) the maximum height attained by the stone above the ground

13. Solution v x (0) = v cos = 19.6 cos30º = 9.8  3 m/s v y (0) = v sin = 19.6 sin30º = 9.8 m/s a x = 0, a y = -9.8 ms –2 (i)y(t) = v y (0)t + a y t 2 –14.7 = 9.8t – 9.8t 2 or 4.9t 2 – 9.8t – 14.7 = 0 or t 2 – 2t – 3 = 0  t = 3 s

14. Solution (ii)R = v x t = = 50.92 m v x (t) = 9.8 3 m/s v y (t) = v y (0) + a y t = 9.8 – 9.8 × 3 or v y = –19.6 m/s

15. Solution Maximum height attained above the ground = Height of B above the point of projection + Height of building

16. Class Exercise - 2 A projectile of mass m is projected with a speed v at an angle  with the horizontal. Which one of the following graphs represents the variation of kinetic energy of the projectile with time? (a) (b) (c) (d)

17. Solution KE is least at the highest point as u sin = 0 Hence answer is (d)

18. Class Exercise - 3 A particle is projected upwards with a velocity of 120 m/s at an angle of 60° to the horizontal. The time after which the particle will be moving upwards at an angle of 45° to the horizontal will be(Take g = 10 m/s 2 )

19. Solution 120 cos60° = v cos45° Hence answer is (a)

20. Class Exercise - 5 If is the equation of a trajectory, then the time of flight will be

21. Solution Comparing the equation with that of the trajectory of a projectile. Hence answer is (a)  tan = 1 or  = 45° and u =

22. Class Exercise - 6 A projectile is thrown at an angle  = 30° with a velocity of 50 m/s in magnitude. Find the time after which the velocity of the projectile makes an angle of 60° with the initial velocity. (Take g = 10 m/s 2 ) Angle between u and v = 2 (Angle of projection) The particle will be at the same horizontal level finally. Solution :

23. Class Exercise - 7 find the time elapsed after the projectile is thrown up. u y = 60, v y = –40 v y = u y – gt –40 = 60 – gt t = 10 s Solution :

24. Instantaneous radius of Curvature ds approximates a circular arc. vtvt v t cos  mg mgcos  x y v  ds

25. Class Exercise - 4 A ball is thrown with a velocity of m/s at an angle of 45° with the horizontal. It just clears two vertical poles of heights 90 cm each. Find the separation between the poles. Solution : Let t be the time after which the ball is at the top of the poles. y(t) = 0.9 m v y (0) = v sin45º Now, y(t) = v y (0)t + a y t 2

26. 0.9 = 7t – (9.8)t 2 4.9t 2 – 7t + 0.9 = 0 Solution On solving, we get Hence the ball is at A after s and at B after s. OP = v x t 1 = 7 × = 1 m OQ = v x t 2 = 7 × = 9 m  PQ = 8 m

27. Class Exercise - 8 What should be the velocity of x such that the objects A, B would collide if both are thrown simultaneously?

28. Solution For the particles to collide, vertical velocity should be the same. So, B should be thrown vertically up with a velocity of 20 m/s.

29. Projectile motion on an inclined plane A particle is thrown horizontally from top of an inclined plane (inclination  with horizontal) with speed . How far from the point of projection will it strike the plane  g O u

30. Projectile motion on an inclined plane Axis ox and oy. Along x axis Along y axis y x g u  

31. Projectile motion on an inclined plane T=0 (particle at O) y x g u  

32. HOME EXERCISE

33. Illustrative Problem What is the product of 2.6  0.5 cm and 2.8  0.5 cm ?

34. Illustrative Problem (b) Divide the vector into two vectors along x’ and y’ which add up the give. (a) Find component of displacement along x’ and y’ axis (c) Are the answer of (a) and (b) are same ? O x’ y’ P 30 o 60 o OP = 50 Km

35. Illustrative Problem A car is traveling at 10m/sec. driver sees a wall at a distance at 150m. he applies brakes at retardation of 1m/sec 2. Find his distance from wall at t = 15 sec ?

36. Illustrative Problem In Searle’s apparatus diameter of the wire was measured 0.05 only screw gauge of least count 0.001 cm. The length of wire was measured 110 cm by meter scale of least count 0.1 cm. an external load of 50 N was applied. the extension in length of wire was measured 0.125 cm by micrometer of least count 0.001 cm. find the maximum possible error in measurement of young’s modules.

37. Y = young modules, A = area of cross section = R 2 where R is radius. Hints

38. Illustrative Problem Figure shows x – t graph of a particle. Find the time t such that the average velocity of the particle during the period 0 to t is zero. x in m t in second 1020 10 O

39. Illustrative Problem A particle starts from a point A and travels along the solid curve shown in figure. find approximately the position B of the particle such that the average velocity between the positions A and B has the same direction as the instantaneous velocity at B. 24 6 x y 2m 4m

40. Thank you