PROJECTILE MOTION 2.4 NCEA LEVEL 2 PHYSICS

2 CONTENTS  Introduction.  Gravity.  Projectile Motion.  Projectile motion calculation summaries.  Projectile motion at an angle.  Video “Pool Projectile Motion”.  Exercises from Rutter.  Homework

3 All projectile motion is caused by the effect of a force. This force we know is GRAVITY. Gravity was first understood and explained by Sir Isaac Newton ( ). Newton proposed that gravity is produced due to the interaction between object (more on that in Level 3). The effect of gravity on an object that is moving is that it causes a parabolic trajectory. Gravity is unlike other forces as it is always attractive. All other forces can be either attractive or repulsive. INTRODUCTION

4 Acceleration due to gravity is constant over the surface of the earth and a few kilometres into the atmosphere. We know it as 9.81ms -2 or rounded to 10ms -2. An object dropped from a height will accelerate straight down at 10ms -2, until it reaches terminal velocity (which we discussed in the motion topic). Using kinematic equations of motion we are able to solve many problems involving gravity, when an object is dropped. v i is always 0 a = 9.81ms -2 d = v i t + ½at 2 d = ½(v i + v f )t v f 2 = v i 2 + 2ad v f = v i + at Using these formula d, t, & v f can be calculated

5 Example 1: A rock (initially stationary) is dropped from a cliff 50m above the sea. a. How far does it fall in 1.0s? b. How far does it fall in 2.0s? c. How long does it take to fall 50m? d = 50m g = 10ms -1 SOLUTION: For a. and b., a velocity time graph can be used. An acceleration of 10ms -2 downwards means that the rock’s velocity increases by 10ms -1 each second v(ms -1 ) t(s) v/t graph Slope 10ms -2

6 a.For 1.0s: (area under graph = distance) d = ½ x 1.0 x 10 d = 5.0m b.For 2.0s: d = ½ x 2.0 x 20 d = 20m c.A kinematic equation canbe used to find the time, t. Since the acceleration of the rock is a = 10ms -2 & v i = 0ms -1 (the rock is stationary), these values can be substituted into the equation: d = v i t + ½at 2 50 = o x t + ½ x 10 x t 2 t 2 = 50/5.0 t =  10 t = 3.2s

7 Example 2: A ball is thrown upwards with an initial speed of 20ms -1 a. Draw a velocity time graph of the motion. b. How far does it rise in 1.0s? c. How long does it take to reach its highest point in its motion? d. How high above the ground is the highest point in its motion? v i = 20ms -1 a = 10ms -1 d/t =?

8 SOLUTION: Slope -10ms -2 v upwards (ms -1 ) t(s) a. The velocity/time graph for the motion is shown alongside. b. Height risen in 1.0s = area under the graph. area = ½ x ( ) x 1.0 area = 15m c. At the highest point, the ball is momentarily stationary (and is about to move downwards). At this moment v = 0 and occurs at t = 2.0s. d. Highest position at t = 2.0s. This height is the area under the graph up to 2.0s. Area = ½ x 20 x 2.0  Height = 20m

9 READ INFORMATION PAGE 32 COMPLETE QUESTIONS FROM RUTTER

10 Projectile motion has two major components: Y component Resultant X component Y component + X component = Resultant Vertical motion (y component), changes due to gravitational acceleration, changing at 9.8 ms Horizontal motion (x component), remains constant as horizontally it is not subjected to gravity. N.B: Air resistance is ignored at this level. PROJECTILE MOTION Any object that moves through the air without its own source of power. E.g bullets, shot puts, netball, water jets and softballs.

11 Vertical component: Gravity acts upon the object changing the velocity at 9.8ms -2. Horizontal component: No gravity acts upon the object. Thus the object moves at a constant velocity. Result of the horizontal & vertical vectors is a parabolic trajectory

12 PROJECTILE MOTION CALCULATIONS 1.Draw a diagram to represent all the information that you have been given. i. Draw the projectile motion arc ii. Draw the resultant or y/x component information. iii. Place on the v f = 0, vertical value. iv. Time lines. v. Range line, calculated using horizontal values using ONLY v = d/t. v = d/t.

13 2.If you have the resultant velocity i.e. the speed and the angle then calculate the i. Y component: hypy adj adj initial y component = cos θ adj/hyp ii. X component: hyp opp opp x initial x component = sin θ opp/hyp

14 3.Finding the maximum height of flight: All of these have to be resolved vertically. i.Find the y component velocity, v i ii.At maximum height the velocity is 0.  v f = 0. iii.Acceleration is constant so a = 9.8 ms -2. iv.Distance is unknown,  d = ?. v.Use the equation v f = v i + 2ad.

15 4.Finding the time of flight: i.Resolve vertically. ii.You know v i (y component) iii.You know v f = 0 as final velocity. iv.You know a = 9.8 ms -2. v.Use v f = v i + at. vi.Rearrange to find ‘t’. This is the time for half of the flight. vii.To find total flight time multiply ‘t’ by 2.

16 5.Finding the range: i.Need to resolve horizontally using ‘v = d / t’ ii.You need to have the full time of flight ‘t’. iii.You need to find the initial horizontal velocity ‘x component’. iv.Rearrange above formula to make ‘d’ the subject. v.Solve.

17 Example 3: A soccer player kicks a ball. The initial components of the velocity of the ball are 10ms -1 vertically and 20ms -1 horizontally. Find the range of the ball. 10ms -1 20ms -1 Range = ? Time = ? Max height v f = 0

18 SOLUTION: At maximum height, the initial vertical speed of 10ms -1 will have reduced to zero: v f = v i + at 0 = 10 + (-10 x t) t = 1.0s Time of flight = twice the time to reach max height T = 1.0 x 2 = 2.0s The ball travels horizontally at a constant speed of 20ms -1 for 2.0s. Thus it has a range of: d = vt = 20 x 2.0  range = 40m

19 Example 4: How high does a softball rise if it is hit with an initial speed of 40ms - 1 at an angle of 40 o to the ground? Max height v f = 0 40 o 40ms -1

20 SOLUTION: Initial vertical speed = 40sin40 o = 25.7ms -1 At the maximum height, the vertical speed will have reduced from 25.7ms -1 to zero. The relevant kinematic equation that does not involve time is: v f 2 = v i 2 + 2ad 0 = (2 x -10 x d)[rearrange] d = -(25.7) 2 / -20 d = 33m

21

22 READ INFORMATION PAGE COMPLETE QUESTIONS FROM RUTTER COMPLETE HOEWORK WORKSHEET ON PROJECTILE MOTION