Projectile Motion. x = v ox t (1)

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1 Projectile Motion Theory Projectile motion is the combination of different motions in the x and y directions. In the x direction, which is taken as parallel to the surface of the earth, the projectile travels with a constant velocity x = v ox t (1) from the moment it is launched until it strikes its target. Of course, this is true only if air resistance can be ignored. In the y direction, which is taken as perpendicular to the surface of the earth, the object experiences a constant acceleration g from the moment it is launched until it strikes its target. y = y o + v oy t gt2 (2) Note that the velocities above are the respective x and y components of the initial vector velocity v o ; θ is the angle relative to the horizontal at which the projectile is launched. v ox = v o cos θ and v 0y = v 0 sin θ. In the classic problem you calculate how far a projectile travels horizontally (its range) given v o and θ. Equation 2 gives the total flight time to the landing height - it yields two times since the projectile is at this height twice along the symmetrical parabolic path. The applicable time in Equation 1 gives the range. Example You are standing on top of a building and throw a golf ball to the right over the side with a speed of 10.0m/s at an angle of 45 above the horizontal. The ball is 20.0m above the ground when it leaves your hand. How long is the ball in the air? How far from the building does it land? Assume no air resistance. Solution Equation 2 y = y o + v oy t gt2 from the experiment gives the total time the ball is in the air. First, we need to define a coordinate system. Let s consider the point at which the ball leaves your hand to be the origin of this coordinate 1

2 system. To the right and up from here are positive x and y respectively. Since the ball leaves at the origin, the initial vertical distance y o = 0m. We then want to find the time at which the final vertical distance y = 20.0m. The only other thing we need is the initial vertical velocity v oy = (10.0m/s)(sin 45 ) = 7.07m/s. Remember that up is positive y, so g = m/s 2. Therefore, y = y o + v oy t gt2 ( 20.0m = 0m + (7.07m/s) t 4.903m/s 2) t 2 ( 0 = 20.0m + (7.07m/s) t 4.903m/s 2) t 2 0 = t 4.903t 2 Solving this quadratic for t yields two times, t 1 = 1.42s and t 2 = 2.87s. The second time here is the one we want. Equation 1 x = v ox t from the experiment gives the horizontal distance traveled. Therefore, x = v ox t = (7.07m/s) (2.87s) = 20.3m and the ball strikes the ground 20.3m to the right of the building. When working through the procedures in this experiment, you should use Equation 2 (only) to find the time of flight. You should then use Equation 1 (only) to find the horizontal distance traveled. Since quadratic equations are involved, you might want to learn how to solve them with your TI calculators - it will save you a great deal of time and aggravation! Apparatus Pasco mini launcher and base, Clamp, Photogate mounting bracket, Photogates, Steel sphere, Ramrod, Lab jack, Meter sticks, Paper, Carbon paper, Masking tape, Plumb bob, Pasco 750, Computer, Capstone software. Procedure Preliminaries Reference Figure 1. The launcher attaches to the base and the photogates attach to the launcher via the photogate mounting bracket (the nut on the front of the bracket slides into the channel under the launcher). Connect the photogate nearest the launcher to digital channel 1 of the 750 and the other to channel 2. Turn on the 750 and computer then open the Projectile Motion activity in Capstone. 2

3 Launcher position and angle are adjusted at the base; do not over-tighten the bolts as they are plastic capped and will break. Just check every few shots to see if the angle has changed. The angle and position of the sphere at launch are displayed on the side of the launcher. The launcher has three ranges - short, medium, and long. These are selectable by how far the sphere is pushed into the launcher; one click for short, two for medium, and three for long. Use a pencil or the provided ramrod to load the sphere in the launcher. Place an object at the far end of the table such as a book bag to stop the sphere from rolling off the table after it lands. Capstone monitors the time it takes the sphere to travel between the two gates and uses it to calculate its initial velocity. To collect data, load the launcher, press the record button in Capstone, then launch the sphere (data collection starts and stops automatically). The velocity will be displayed and the process repeated. Tape a sheet of paper where the sphere lands and lay a piece of carbon paper over it to mark the location. Figure 1: Arrangement of components. Firing Up to a Higher Level 1. Clamp the base at the middle of one of the ends of the table so that the sphere will travel down the center of its length when launched. Select a launch angle 35 < θ < 55 and position such that the sphere is over the table at launch. Adjust the lab jack to a height 10cm < h < 20cm. 2. Measure y o and y relative to the table top. 3. Launch some test shots at short range and position the jack such that the sphere strikes the top of the jack on its way down. 4. Launch the sphere five times similarly; note each v o and measure the horizontal distance traveled. 3

4 Firing Up to a Lower Level This procedure is similar to the previous with the following exceptions: 1. Move the base to the edge of the table so that the sphere will land on the floor once launched. 2. Use the plumb bob to locate the point on the floor at which the sphere leaves the launcher. 3. You may use short or medium range; if medium, make sure you launch to a clear area. 4. y o and y are measured relative to the floor. Firing Down to a Lower Level This procedure is similar to the previous with the only exception being a launch angle below the horizontal. Analysis Note: As always, there is uncertainty in all the quantities measured - as well as other quantities you have and will calculate. However, we will dispense with these in this experiment. Why? The largest uncertainty should be in v o ; i.e., we could ignore the uncertainties in the remaining values. However, even this uncertainty is relatively small and it is used to calculate time which involves a quadratic polynomial. Firing Up to a Higher Level 1. Calculate v o and use this for subsequent calculations involving v o. 2. Substitute the corresponding values into Equation 2, write it in the form At 2 + Bt + C = 0, and solve it for the two times the sphere is at the jack height. 3. Use the appropriate time in Equation 1 to calculate how far the sphere should travel. Compare (percent error or percent difference) this distance to your measured x. Firing Up to a Lower Level Same analysis as in the previous (though your two times are when the sphere is at floor level). Firing Down to a Lower Level Same analysis as in the previous. 4

5 Pre-Lab: Projectile Motion Name Section Answer the questions at the bottom of this sheet, below the line - continue on the back if you need more room. Any calculations should be shown in full. 1. A projectile is launched at 10.0m/s at a height of 5.000m above the ground and an angle of 20 below the horizontal. What are the horizontal and vertical components of this velocity? 2. You launch the projectile in Question 1 as described; calculate the two times the projectile is at ground level with Equation Using the appropriate time from Question 2, calculate the horizontal distance the projectile will travel before striking the ground. 4. When you solve Equation 2 for time, is it theoretically possible to obtain a negative value? Why or why not? 5