Motion in 2 Dimensions

1. Motion in 2 Dimensions

2. Definitions • projectile • any object thrown or otherwise projected into the air • trajectory • the parabolic path of a projectile

3. In projectile motion, the horizontal and the vertical components of the motion are treated separately. A projectile moves both horizontally and vertically. Its horizontal motion is constant. Its vertical motion is affected by the acceleration due to gravity. The only variable shared by both types of motion is time. Every point on the trajectory is the vector sum of the horizontal and the vertical components of the velocity.

4. An object projected horizontally (projected perfectly parallel to the surface) will reach the ground in the same time as an object dropped vertically. Since speed at any point in a trajectory is the vector sum of the horizontal and vertical velocity components at that point, the projected object will have a greater speed when it strikes. • The maximum range for a given initial velocity is obtained when the angle of projection is 45°.

5. Equations that are used to describe the horizontal motion: horizontal motion (constant velocity): • x = xo + voxt + ½ axt2 • Simplified to • x = voxt Note: where vox is the horizontal velocity component and x is the horizontal distance (range)

6. Equations that are used to describe the vertical motion: vertical motion (acceleration): • y= yo + voyt + ½ ayt2 • vy = voy + ayt • vy2 = voy2 + 2ay(Dy) • Note: where y is the vertical distance, voy is the initial vertical velocity, and vy is the final vertical velocity

7. Speed • The speed of an object at any point on the trajectory can be found by calculating the horizontal and vertical velocity components at that point. The speed is the vector sum of the components at that point. This is a very common "trick" that arises on the AP Physics B exam. You are asked to calculate speed, acceleration, etc. It is very easy to forget that it is the resultant of the horizontal and the vertical components of speed, acceleration, etc. • For example, assume you are asked to calculate the speed of an object which rolls off a vertical cliff with horizontal velocity vx. You must calculate the final vertical velocity (vy)of the object the instant before the object strikes. The speed is the square root of the sum of (vx)2 and (vy)2.

8. Directions • It is very important to consistently define directions in projectile motion. If the acceleration due to gravity is defined to be negative, then all velocities in the "down" direction are also negative. If displacement is measured from the ground up, it is positive. If it is measured from the "top" down, it is negative.

9. AP assignment of an origin • In AP problems, it helps to assign an "origin." For example, an object is thrown upward from the top of a cliff with speed v and at an angle q. Call this point the origin. It is convenient to set the origin where t=0. The initial vertical velocity will thus be positive. The height of the cliff would be negative because it is measured from your origin downward in the -y direction.

10. Advanced calculations The general equation of motion, y = yo + voyt + ½ ayt2, can be easily used to calculate vertical displacement and/or time at any point in a trajectory. First, find the initial vertical velocity component. • If horizontal distance is known: Use the horizontal velocity component to calculate time in the air to that point in the trajectory. Substitute that value for time in y = yo + voyt + ½ ayt2 and solve for vertical displacement, knowing the initial vertical velocity component. • If vertical displacement is known (remember to assign your origin system and use the appropriate positive and negative signs for vertical displacement and initial vertical velocity): Use y = yo + voyt + ½ ayt2 to solve for time to that point in the trajectory knowing the initial vertical velocity component.

11. Advanced calculations(con’t) • Calculate the total time the object is in the air by using y = yo + voyt + ½ ayt2 and the initial vertical velocity component. Remember, to set your origin and use the appropriate positive and negative signs for the initial vertical velocity component and for vertical displacement. • Suppose an object is launched from the ground with speed v and at an angle q. You can use d y = yo + voyt + ½ ayt2 and the initial vertical velocity component to find the total time in the air. Remember, when it hits the ground, the vertical displacement would be zero.

12. We will be working two types of problems The first type of problem involves an object that is thrown or projected at the top (or apex) of its trajectory (at point B in the picture). At this point, the vertical velocity is zero. Since it starts at this point, we let voy = zero. It accelerates vertically downward at a rate of -9.8 m/s2. We usually use the first acceleration formula listed to find the time the projectile is in the air. The time that it travels vertically is the same as it travels horizontally.

13. We can use this time to determine its horizontal range. Its vertical displacement is given by distance BD and it is negative. Its horizontal range is given by distance DC. • vertical: vy = v sin q • horizontal: vox= v cos q In this case, q = 0°. Thus, cos q = 1 and sin q = 0 which makes the initial vertical velocity component equal to zero and the horizontal velocity component equal the horizontally projected velocity.

14. The second type of problem involes an object that is projected from the ground (point A) and whose path is parabolic. At this point, the speed at which it is projected, v, is the vector sum of its horizontal and its vertical components. At point A, its initial vertical velocity component and its horizontal velocity component are given by: vertical: vo = v sin q horizontal: vo = v cos q

15. At point B, its vertical velocity component is zero. The acceleration at all points is -9.8 m/s2. At point B, the object is accelerating (even though its vertical velocity is zero), because its direction is changing. At all points in the trajectory, the horizontal velocity component remains the same (in the absence of air friction). We will work these problems by finding total time in the air using the second acceleration formula. vat point C is numerically equal to vo at point A but of the opposite sign.

16. Remember, velocity is a vector quantity! Once we have total time, we can use d = v t to find the horizontal range, knowing that this uses the horizontal velocity componet. We can use the last acceleration formula to predict the maximum vertical displacement (working from points B and D).

17. A formula can be derived for the horizontal range: • Range = (v2 sin 2q) / g • where v is the projectile velocity, q is the projectile angle, and g is the acceleration due to gravity. Advanced calculations • The general equation of motion, d = vot + ½ a t2, can be easily used to calculate vertical displacement and/or time at any point in a trajectory. You can also calculate the vertical velocity component at any point knowing the initial vertical velocity component and the time to that point. Remember to set your origin and use the appropriate positive and negative signs. You can then use vf = vi + at to find the vertical velocity component at that point.