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1 Instructions To run the slideshow: Click: view full screen mode, or press Ctrl +L. Left click advances one slide, right click returns to previous slide. To exit the slideshow press the Esc key.
2 Monkey and Hunter  Projectile Motion KS5
3 Monkey and Hunter A hunter is trying to shoot a monkey hanging from a tree branch. (With a mild sedative) The Hunter knows that in order to avoid the bullet the monkey will let go of the branch the second the trigger is pulled. Where should the Hunter aim in order to hit the monkey?
4 Projectiles. In this problem both the falling monkey and the fired bullet can be considered as projectiles. The exam board definition of a projectile is: Any object thrown in the Earth s gravitational field.
5 More on Projectiles: Dictionary Definition: n. A fired, thrown, or otherwise propelled object, such as a bullet, having no capacity for selfpropulsion. A projectile is an object on which the primary force acting is the force due to gravity. In most exam questions air resistance is considered negligible and so we can treat any object moving through the air on Earth as a projectile. (provided it is not self propelled).
6 A simple projectile: The monkey in this problem is a very simple projectile. 1. Its motion is only in the vertical plane. 2. It travels in only one direction. Given that the monkey is experiencing a constant acceleration (g) we can apply the equations of motion to model the path of the monkey.
7 The monkey We know that the monkey starts from rest: u=0ms 1 We know the acceleration of the monkey: a=9.81ms 2 s = (u + v) 2 t s = ut at2 v 2 = u 2 + 2as Given any other variable we can now use the equations of motion in the usual way.
8 Example The monkey is hanging 4m from the ground. s = (u + v) 2 t Calculate how long it will take for it to land on the floor. s = ut at2 v 2 = u 2 + 2as
9 Solution The monkey is hanging 4m from the ground. Calculate how long it will take for it to land on the floor. o Select the most suitable equation: s = ut at2 s = (u + v) 2 t o Rearrange and substitute in the values: s = ut at2 v 2 = u 2 + 2as s = 1 2 at2 2s a = t t = 0.9s
10 4.5 In detail The graphic on the left shows the position of the monkey plotted at 0.2s intervals It is clear that the distance travelled in each 0.2s interval increases as the monkey falls to the ground This result should not be surprising as the acceleration of the monkey is constant. 0
11 The hunter s bullet The bullet is a slightly more complicated problem. For ease we will consider a bullet fired horizontally. We now have an object that is travelling in both a horizontal and vertical direction.
12 Horizontally s H Looking at the horizontal displacement s H we can see that for each time interval the displacement is constant. This implies a constant velocity and zero acceleration. (this is only true for negligible air resistance) Use the equation v = s t 0
13 Vertically s v Looking at the vertical displacement s v we can see that for each time interval the displacement is increasing. This implies an increasing velocity as would be expected with an acceleration (g). Use the equations of motion.
14 For any projectile: If we resolve the initial velocity into horizontal and vertical components we can treat the two independently. (This is true of any perpendicular components of a vector) Horizontally u = Horizontal component of initial velocity. a = 0 v = u (as acceleration is 0) Use v = s t Vertically u = Vertical component of initial velocity. a = 9.81ms 2 Use the equations of motion
15 Monkey and hunter. Assuming the bullet is fired horizontally at the monkey and calculating vertical displacement for each: Monkey s = ut at2 u = 0 a = 9.81ms 2 s = 4.9t 2 Hunter s = ut at2 u = 0 a = 9.81ms 2 s = 4.9t 2 Because for the bullet the vertical motion is independent of the horizontal motion, we have the same vertical solution for both the bullet and the monkey.
16 Where should the hunter aim? Given the previous information. Where should the hunter aim to ensure that he hits the monkey?
17 Monkey and Hunter The monkey is held aloft by a small electromagnet. As the bullet is fired a switch is triggered which releases the monkey.
18 Monkey and Hunter Regardless of height, initial velocity or distance between them, if the hunter aims directly at the monkey, and the monkey lets go at the time the bullet is fired. The hunter will always hit the monkey. The only exception to this is if the horizontal velocity of the bullet is small enough to allow the monkey to reach the ground. In this case the bullet will hit the ground harmlessly between the monkey and the hunter.
19 Graphically Plotting the vertical motion of both the bullet and the monkey on the same axis shows clearly that the bullet will always hit the monkey
20 More on projectiles A classic projectile is not fired horizontally but at an angle to the ground:
21 The path that the projectile takes is called its trajectory. It follows a mathematical shape called a parabola. Trajectory In this instance the projectile is moving both horizontally and vertically. It is also travelling initially in the positive x direction and then in the negative x direction. It is important now to pay particular attention to sign conventions.
22 Example 1 Another simple projectile question. A ball is thrown vertically into the air with an initial velocity of 20ms 1 Calculate the maximum height of the ball.
23 Solution 1 Another simple projectile question. A ball is thrown vertically into the air with an initial velocity of 20ms 1 Calculate the maximum height of the ball. v 2 = u 2 + 2as 0 = s u = 20ms 1 v = 0 a = 9.81ms 2 s = 0 = s s = = 20.4m
24 Example 2 A cannon ball is launched with an initial velocity of 50ms 1 at an angle of 30 0 to the horizontal. Calculate the range of the cannonball. (The range is the maximum horizontal distance travelled) (Hint: time t will be the same for the horizontal and vertical motion)
25 Solution 2 First resolve the velocity into horizontal and vertical components: u v = 50sin30 u h = 50cos30 Taking vertical motion first: s = ut at2 0 = ut at2 s = 0 (ball will return to the ground) u = 50sin30 a = 9.81ms 2 t = ut = 1 2 at2 2u = t = 5.1s 9.81
26 Solution 2 Cont. Taking motion horizontally: s = ut s = 50cos s = u = 50cos30ms 1 a = 0 (horizontal a = 0) t = 5.1s s = 220m
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