Summary Physics grade 12 physics ch21 motion in two dimensions

Motion in Two Dimensions - Grade 12 21.1 Introduction In Chapter 3, we studied motion in one dimension and briefly looked at vertical motion. In this chapter we will discuss vertical motion and also look at motion in two dimensions. In Chapter 12, we studied the conservation of momentum and looked at applications in one dimension. In this chapter we will look at momentum in two dimensions. 21.2 Vertical Projectile Motion In Chapter 4, we studied the motion of objects in free fall and we saw that an object in free fall falls with gravitational acceleration g. Now we can consider the motion of objects that are thrown upwards and then fall back to the Earth. We call this projectile motion and we will only consider the situation where the object is thrown straight upwards and then falls straight downwards - this means that there is no horizontal displacement of the object, only a vertical displacement. 21.2.1 Motion in a Gravitational Field When an object is in a gravitational field, it always accelerates downwards with a constant acceleration g whether the object is moving upward or downward. This is shown in Figure 21.1. Important: Projectiles moving upwards or downwards always accelerate downwards with a constant acceleration g. bobject moving upwards g bobject moving downwardsg Figure 21.1: Objects moving upwards or downwards, always accelerate downwards. This means that if an object is moving upwards, it decreases until it stops (vf = 0 m·s−1). This is the maximum height that the object reaches, because after this, the object starts to fall. Important: Projectiles have zero velocity at their greatest height. 463 21.2 CHAPTER 21. MOTION IN TWO DIMENSIONS - GRADE 12 Consider an object thrown upwards from a vertical height ho. We have seen that the object will travel upwards with decreasing velocity until it stops, at which point it starts falling. The time that it takes for the object to fall down to height ho is the same as the time taken for the object to reach its maximum height from height ho. initial height h0 maximum height b (a) time = 0 s b (b) time = tm b (c) time = 2tm Figure 21.2: (a) An object is thrown upwards from height h0. (b) After time tm, the object reaches its maximum height, and starts to fall. (c) After a time 2tm the object returns to height h0. Important: Projectiles take the same the time to reach their greatest height from the point of upward launch as the time they take to fall back to the point of launch. 21.2.2 Equations of Motion The equations of motion that were used in Chapter 4 to describe free fall can be used for projectile motion. These equations are the same as those equations that were derived in Chapter 3, but with a = g. We use g = 9,8 m · s−2 for our calculations. vi = initial velocity (m·s−1) at t = 0 s vf = final velocity (m·s−1) at time t ∆x = height above ground (m) t = time (s) ∆t = time interval (s) g = acceleration due to gravity (m·s−2) vf = vi + gt (21.1) ∆x = (vi + vf ) 2 t (21.2) ∆x = vit + 1 2 gt2 (21.3) v2 f = v2 i + 2g∆x (21.4) Worked Example 132: Projectile motion Question: A ball is thrown upwards with an initial velocity of 10 m·s−1. 1. Determine the maximum height reached above the thrower’s hand. 2. Determine the time it takes the ball to reach its maximum height. Answer 464 CHAPTER 21. MOTION IN TWO DIMENSIONS - GRADE 12 21.2 Step 1 : Identify what is required and what is given We are required to determine the maximum height reached by the ball and how long it takes to reach this height. We are given the initial velocity vi = 10 m·s−1and the acceleration due to gravity g = 9,8 m·s−2. Step 2 : Determine how to approach the problem Choose down as positive. We know that at the maximum height the velocity of the ball is 0 m·s−1. We therefore have the following: • vi = −10 m · s−1 (it is negative because we chose upwards as positive) • vf = 0 m · s−1 • g = +9,8 m · s−2 Step 3 : Identify the appropriate equation to determine the height. We can use: v2 f = v2 i + 2g∆x to solve for the height. Step 4 : Substitute the values in and find the height. v2 f = v2 i + 2g∆x (0)2 = (−10)2 + (2)(9,8)(∆x) −100 = 19,6∆x ∆x = 5,102...m The value for the displacement will be negative because the displacement is upwards and we have chosen downward as positive (and upward as negative). The height will be a positive number, h = 5.10m. Step 5 : Identify the appropriate equation to determine the time. We can use: vf = vi + gt to solve for the time. Step 6 : Substitute the values in and find the time...