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1© Manhattan Press (H.K.) Ltd. Monkey and hunter experiment Body projected horizontally under gravity Body projected horizontally under gravity Body projected.

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Presentation on theme: "1© Manhattan Press (H.K.) Ltd. Monkey and hunter experiment Body projected horizontally under gravity Body projected horizontally under gravity Body projected."— Presentation transcript:

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2 1© Manhattan Press (H.K.) Ltd. Monkey and hunter experiment Body projected horizontally under gravity Body projected horizontally under gravity Body projected at an angle under gravity Body projected at an angle  under gravity 4.1 Independence of horizontal and vertical motions

3 2 © Manhattan Press (H.K.) Ltd. What is “projectile motion”? Kicking a football 4.1 Independence of horizontal and vertical motions (SB p. 144) The football will be projected along two-dimensional path

4 3 © Manhattan Press (H.K.) Ltd. 4.1 Independence of horizontal and vertical motions (SB p. 144) Independence of horizontal and vertical motions Assume air resistance is negligible. During the flight, (1)the horizontal velocity remains constant (2)the vertical velocity is subjected to a constant acceleration due to gravity

5 4 © Manhattan Press (H.K.) Ltd. 4.1 Independence of horizontal and vertical motions (SB p. 145) Monkey and hunter experiment bullet leaves the mouth of the toy gun1 breaks the aluminium strip and the circuit of the electromagnet 2 iron monkey falls vertically downwards 3 the bullet always hits the monkey 4

6 5 © Manhattan Press (H.K.) Ltd. 4.1 Independence of horizontal and vertical motions (SB p. 145) Monkey and hunter experiment The bullet and the monkey start with zero vertical downward velocity They fall under the same acceleration due to gravity their vertical motions are identical Their vertical positions are the same at any instant of time Why does the bullet always hit the monkey?

7 6 © Manhattan Press (H.K.) Ltd. 4.1 Independence of horizontal and vertical motions (SB p. 145) Monkey and hunter experiment The horizontal and vertical components of a projectile motion are independent of each other. We can conclude that:

8 7 © Manhattan Press (H.K.) Ltd. 4.1 Independence of horizontal and vertical motions (SB p. 146) Body projected horizontally under gravity a marble is projected horizontally off the edge of a table travels with a uniform velocity in the horizontal direction and accelerates uniformly downwards due to gravity the horizontal velocity remains constant

9 8 © Manhattan Press (H.K.) Ltd. 4.1 Independence of horizontal and vertical motions (SB p. 146) Body projected horizontally under gravity horizontal motion vertical motion They are independent of each other

10 9 © Manhattan Press (H.K.) Ltd. 4.1 Independence of horizontal and vertical motions (SB p. 146) Body projected horizontally under gravity Horizontal motion (with constant velocity) Vertical motion (with constant acceleration a = -g) Velocity Displacement v x = u x = constant v y = -gt, u y = 0 (-ve means moving downwards) s x = v x t = u x t s y = - gt 2

11 10 © Manhattan Press (H.K.) Ltd. 4.1 Independence of horizontal and vertical motions (SB p. 147) Body projected horizontally under gravity 1.Instantaneous velocity (v) of the projectile: 2.Direction of motion at certain instant:

12 11 © Manhattan Press (H.K.) Ltd. 4.1 Independence of horizontal and vertical motions (SB p. 147) Body projected horizontally under gravity 3. Time of flight t f :

13 12 © Manhattan Press (H.K.) Ltd. 4.1 Independence of horizontal and vertical motions (SB p. 147) Body projected horizontally under gravity 4.The total horizontal displacement travelled by the object is called the range (R).

14 13 © Manhattan Press (H.K.) Ltd. 4.1 Independence of horizontal and vertical motions (SB p. 147) Body projected horizontally under gravity Trajectory of the projectile: s x = u x t and

15 14 © Manhattan Press (H.K.) Ltd. Let s x be x and s y be y. 4.1 Independence of horizontal and vertical motions (SB p. 147) Body projected horizontally under gravity The equation is of the parabolic form y = kx 2 Trajectory of the projectile is a parabola Go to Example 1 Example 1 Go to Example 2 Example 2

16 15 © Manhattan Press (H.K.) Ltd. 4.1 Independence of horizontal and vertical motions (SB p. 152) Body projected at an angle  under gravity Initial vertical velocity: u y = u sin  Initial horizontal velocity: u x = u cos  the object is projected with a velocity u at an angle  to the horizontal

17 16 © Manhattan Press (H.K.) Ltd. 4.1 Independence of horizontal and vertical motions (SB p. 152) The vertical motion of the object undergoes constant acceleration a = –g Vertical component of velocity (v y ) and the vertical displacement (y) at time t : Body projected at an angle  under gravity

18 17 © Manhattan Press (H.K.) Ltd. 4.1 Independence of horizontal and vertical motions (SB p. 152) The horizontal motion is uniform and the object moves with constant velocity Horizontal velocity component (v x ) and the horizontal displacement (x) at time t: Body projected at an angle  under gravity

19 18 © Manhattan Press (H.K.) Ltd. Instantaneous velocity (v): The direction of motion at certain instant: 4.1 Independence of horizontal and vertical motions (SB p. 152) Body projected at an angle  under gravity

20 19 © Manhattan Press (H.K.) Ltd. 4.1 Independence of horizontal and vertical motions (SB p. 153) At B, the object reaches its maximum height H with v y = 0, Body projected at an angle  under gravity

21 20 © Manhattan Press (H.K.) Ltd. 4.1 Independence of horizontal and vertical motions (SB p. 153) At D, the time of flight of the object (t f ) Body projected at an angle  under gravity

22 21 © Manhattan Press (H.K.) Ltd. 4.1 Independence of horizontal and vertical motions (SB p. 153) Horizontal displacement OD of the projectile is the range R Body projected at an angle  under gravity

23 22 © Manhattan Press (H.K.) Ltd. By combining: 4.1 Independence of horizontal and vertical motions (SB p. 153) The equation is of the parabolic form y = ax – bx 2 trajectory of the projectile is a parabola where 0 o <  < 90 o Body projected at an angle  under gravity

24 23 © Manhattan Press (H.K.) Ltd. 4.1 Independence of horizontal and vertical motions (SB p. 154) occurs when sin2  = 1 or 2  = 90 o  = 45 o  maximum range is obtained if the object is projected at an angle 45 o to the horizontal Note: 1.The vertical and horizontal motions of a projectile motion are independent of each other 2. The trajectory of the projectile is parabolic 3. The maximum value of R is Body projected at an angle  under gravity

25 24 © Manhattan Press (H.K.) Ltd. 4.1 Independence of horizontal and vertical motions (SB p. 154) Note: 4.  sin2  = sin2(90 o –  ) Maximum range at  = 45 o Range at 60 o = Range at 30 o two angles of projection for a given range R with a specific speed, except for  = 45 o Body projected at an angle  under gravity

26 25 © Manhattan Press (H.K.) Ltd. 4.1 Independence of horizontal and vertical motions (SB p. 154) Note: 5. In real situation, air resistance reduces the speed and the range of the projectile Go to More to Know 1 More to Know 1 Go to Example 3 Example 3 Body projected at an angle  under gravity

27 26 © Manhattan Press (H.K.) Ltd. End

28 27 © Manhattan Press (H.K.) Ltd. 4.1 Independence of horizontal and vertical motions (SB p. 148) Q: Q:An aeroplane, flying in a straight line at a constant height of 500 m with a speed of 200 m s –1, drops an object. The object takes a time t to reach the ground and travels a horizontal distance d in doing so. Taking g as 10 m s –2 and ignoring air resistance, what are the values of t and d ? Solution

29 28 © Manhattan Press (H.K.) Ltd. 4.1 Independence of horizontal and vertical motions (SB p. 148) Solution: For the horizontal motion: Distance (d) = 200  t = 200  10 = m When the object is released from the aeroplane, Horizontal component of velocity = Velocity of the aeroplane = 200 m s –1 For the vertical motion of the object: Initial velocity = 0, acceleration = g = 10 m s –2, displacement = 500 m Using 500 = 0 +  10  t 2 t 2 = 100 t = 10 s Return to Text

30 29 © Manhattan Press (H.K.) Ltd. Q: Q:When a rifle is fired horizontally at a target P on a screen at a range of 25 m, the bullet strikes the screen at a point 5.0 mm below P. The screen is now moved to a distance of 50 m from the rifle and the rifle again is fired horizontally at P in its new position. Assuming that air resistance may be neglected, what is the new distance below P at which the screen would now be struck? Solution 4.1 Independence of horizontal and vertical motions (SB p. 149)

31 30 © Manhattan Press (H.K.) Ltd. 4.1 Independence of horizontal and vertical motions (SB p. 149) Solution: Solution: Let v = velocity of the bullet when it leaves the rifle.  Time taken for the bullet to travel through a horizontal distance of 25 m (t 1 ) = Consider vertical motion of the bullet, initial velocity (u) = 0, acceleration (a) = g, displacement (s) = 5.0  10 –3 m, time (t 1 ) = Using

32 31 © Manhattan Press (H.K.) Ltd. 4.1 Independence of horizontal and vertical motions (SB p. 149) Solution (cont’d): Solution (cont’d): With the screen at 50 m from the rifle, time of travel (t 2 ) = Using New distance below P (h) = (2)  (1) Return to Text

33 32 © Manhattan Press (H.K.) Ltd. 4.1 Independence of horizontal and vertical motions (SB p. 154) So far, air resistance has been neglected for projectile motions. In real situations, the trajectory of the projectile is influenced by air resistance in the extent that depends on the mass, shape and size of the objects. Return to Text

34 33 © Manhattan Press (H.K.) Ltd. Q: Q: An aeroplane flies at a height h with a constant horizontal velocity u so as to fly over a cannon. When the aeroplane is directly over the cannon, a shell was fired to hit the aeroplane. Neglecting air resistance, what is the minimum speed of the shell in order to hit the aeroplane? (g = acceleration due to gravity) 4.1 Independence of horizontal and vertical motions (SB p. 154) Solution

35 34 © Manhattan Press (H.K.) Ltd. Solution: Solution: Letv = minimum speed of shell,  = angle of projection. The minimum speed is when the shell will hit the aeroplane at the maximum height of the shell trajectory. Since the shell is fired when the aeroplane is directly above it, the horizontal component of velocity of the shell must be the same as the velocity of the aeroplane, i.e. same u. 4.1 Independence of horizontal and vertical motions (SB p. 154)

36 35 © Manhattan Press (H.K.) Ltd. Solution (cont’d): Solution (cont’d): Using the equation, Horizontal component of velocity of the shell, 4.1 Independence of horizontal and vertical motions (SB p. 154) Return to Text


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