2-D Kinematics. In general, we have the following 8 equations (4 per dimension): Notes Page 1 of 7

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Charlene Marybeth Campbell
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1 2-D Kinematics The problem we run into with 1-D kinematics, is that well it s one dimensional. We will now study kinematics in two dimensions. Obviously the real world happens in three dimensions, but that s for a university level mechanics course. Perhaps the most important concept to understand while studying 2-D Kinematics is that the horizontal and vertical components are completely independent of each other! In other words, the motion in one direction has no effect on the other! We ll demonstrate this in class because it may seems counterintuitive to you. We will use 2-D kinematics to mostly analyze projectile motion one horizontal and one vertical dimension. We will deal with what is called simple projectile motion. Simple projectile motion operates under one very important assumption: The only acceleration of the object is due to Earth s gravity. This tells us that a y = g always! It also tells us that a x = 0 m s 2, which implies that Δv x = 0 or v 0x = v x (In words: the x-velocity is constant). The simple projectile motion assumption is most accurate when objects are dense, have a small cross-sectional area, and have a small displacement. In general, we have the following 8 equations (4 per dimension): x-direction y-direction v x = a x t + v 0x v y = a y t + v 0y v 2 2 x = 2a x Δx + v 0x v 2 2 y = 2a y Δy + v 0y x = 1 2 a xt 2 + v 0x t + x 0 y = 1 2 a yt 2 + v 0y t + y 0 x = 1 2 (v x + v 0x )t + x 0 y = 1 2 (v y + v 0y )t + y 0 (Note: Δx = v t x x 0 = ( v+v 0 2 ) t x = 1 2 (v + v o)t + x 0 ) Notes Page 1 of 7

2 If we apply the following assumptions of simple projectile motion the above 8 equations will simplify: Assumptions: o Air resistance is negligible a x = 0 m s 2 v 0x = v x = a constant. o This also means a y = g = 9.80 m at all times (if up is positive ) s2 x-direction x = v 0x t + x 0 v 0x = v x y-direction v y = gt + v 0y v 2 2 y = 2gΔy + v 0y y = 1 2 gt2 + v 0y t + y 0 y = 1 2 (v y + v 0y )t + y 0 If your reference angle θ is relative to the positive x-axis, you can also make the following substitutions: v 0x = v 0 cosθ v 0y = v 0 sinθ θ = tan 1 ( v y v x ) Let s construct a motion diagram for the trajectory of a simple projectile: Notes Page 2 of 7

3 Special Case Shortcuts [Completely optional but useful] 1. If a projectile is launched with a velocity v 0 and an angle θ above the horizontal and y 0 = y f (or Δy = 0), then find an equation for Δx. In this case, Δx is sometimes called the range and the equation you ll find is often called the range equation. This equation can be helpful to know as a time saver. 2. If v 0y = 0 m, then find the time of flight of the projectile. I usually call this type of s problem a cliff problem because v oy = 0 m if you drive straight off a flat cliff (which s comes up a lot in basic physics for some reason!) Notes Page 3 of 7

4 Problems 1. Consider the three scenarios below. For each scenario, the projectile is launched from the same initial height, at the same speed, and at the same initial angle. The projectiles all reach their maximum height before landing. Rank the projectiles from greatest to least flight time and from greatest to least final speed. Justify your answers. I. The projectile is launched so that y final = y 0. II. The projectile is launched so that y final > y 0 III. The projectile is launched so that y final < y A football is kicked from a ground-level tee at an angle of with an initial velocity of 20.0 m/s. Calculate the following: a) The maximum height the ball reaches b) The total time the ball is in the air c) How far away from the kicker the ball hits the ground d) The velocity vector at the maximum height e) The acceleration vector at maximum height. Notes Page 4 of 7

5 3. A soccer player kicks a rock horizontally off a 40.0-m-high cliff into a pool of water. If the player hears the sound of the splash 3.00s later, what was the initial speed given to the rock? Assume the speed of sound in air to be 343 m/s. 4. A projectile is launched with an initial speed of 60.0 m/s at an angle of above the horizontal. The projectile lands on a hillside 4.00s later. Neglect air friction. (a) What is the projectile s velocity at the highest point of its trajectory? (b) What is the displacement of the rocket? Notes Page 5 of 7

6 5. A daredevil is shot out of a canon at 45.0 o to the horizontal with an initial speed of 25.0 m/s. A net is positioned a horizontal distance of 50.0m from the canon. At what height should the net be placed in order to catch the daredevil? 6. Two projectiles are launched simultaneously from the same height and at the same speed, v 0. One projectile is launched at an angle θ above the horizontal and another is launched at the same angle θ below the horizontal. What is the difference in the two projectiles final horizontal displacements? Answer in terms of v 0, θ, and g.. Notes Page 6 of 7

7 7. Consider the figure. How fast must the basketball player shoot the ball so that it goes through the hoop without touching the backboard? Notes Page 7 of 7

Chapter 2. Kinematics in One Dimension. continued

Chapter 2. Kinematics in One Dimension. continued Chapter 2 Kinematics in One Dimension continued 2.6 Freely Falling Bodies Example 10 A Falling Stone A stone is dropped from the top of a tall building. After 3.00s of free fall, what is the displacement