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Simple Projectile Motion

Background

● Background Overview

● Projectile Motion

Projectile motion is the motion of an object thrown or projected into the air, subject only to acceleration as a result of gravity. The applications of projectile motion in physics and engineering are numerous. Some examples include meteors as they enter Earth’s atmosphere, water in a water fountain, and the motion of any ball in sports. Such objects are called projectiles, and their path is called a trajectory.

We can represent a projectile’s motion through kinematics which utilize its position, time, velocity, and acceleration. The kinematic equations we will use during this lab assume both constant acceleration and that the effects of air resistance are negligible (generalized in the next four equations with xx representing position along any given dimension):

v=v0+atv = v_{0} + at
x=x0+v0t+12at2x = x_{0} + v_{0}t + \frac{1}{2}at^{2}
v2=v02+2a(xx0)v^{2} = v_{0}^{2} + 2a(x - x_{0})
x=x0+12(v0+v)tx = x_{0} + \frac{1}{2}(v_{0} + v)t

Today, we will analyze two-dimensional trajectories, where xx will represent the horizontal direction, and yy the vertical. Gravity (which pulls objects down towards the center of the Earth) will be assumed to be the only source of acceleration acting on our projectiles; as such, the acceleration of a projectile is ay=ga_{y} = -g and ax=0a_{x} = 0. Since horizontal acceleration terms are zero, we can represent today’s horizontal motion by:

v0x=vx=constant velocity,   x=x0+v0xtv_{0x} = v_{x} = \text{constant velocity},~~~ x = x_{0} + v_{0x}t

Then, depending on the information we have available to us, we can represent today’s vertical motion by:

vy=v0ygtv_{y} = v_{0y} - gt
y=y0+v0yt12gt2y = y_{0} + v_{0y}t - \frac{1}{2}gt^{2}
vy2=v0y22g(yy0)v_{y}^{2} = v_{0y}^{2} - 2g(y - y_{0})
y=y0+12(v0y+vy)ty = y_{0} + \frac{1}{2}(v_{0y} + v_{y})t

● Experimental Summary

The setup today will involve a marble launcher that can slide into a large holder at different initial heights y0y_{0} (and later for the third case, different angle(s), see Figure 1). Once the ball is launched, it will begin a two-dimensional trajectory accelerated solely by gravity in the yy direction.

For the first case, you will measure distances traveled in both xx and yy directions (see Figure 2). For the second and third cases, you will calculate the theoretical distance (based on your results from Case 1) in the xdirectionx\,\text{direction}, mark on the landing paper where your estimated landing zone would be, and then test with your experimental trials (see Figure 3).

Stars denote projectile height. Left) Position of launcher for lower height in Case 1. Right) Position of launcher for higher height in Case 2. Right) Position of launcher for angled launches in Case 3 where the ball’s initial position is at the same higher height position.

Figure 1:Stars denote projectile height. Left) Position of launcher for lower height in Case 1. Right) Position of launcher for higher height in Case 2. Right) Position of launcher for angled launches in Case 3 where the ball’s initial position is at the same higher height position.

Example of Case 1 data aquisition.

Figure 2:Example of Case 1 data aquisition.

Example of Case 2 and 3 estimation and data aquisition.

Figure 3:Example of Case 2 and 3 estimation and data aquisition.

○ Zero-Angle Trajectories (Background for Cases 1 & 2)

The distance traveled in the in the xx direction Δx\Delta x will be measured from the center of the ball in the uncocked position (initial position x0x_0) to the average landing position on the floor (final position xx) after the given number of trials. Note: Δx=xx0\Delta x = x - x_0, however since initial position x0=0mx_0 = 0\,\text{m} for all cases throughout the lab today, Δx=x\Delta x = x.

The ball will mark up white printer paper with black carbon paper; we will note each landing point (e.g. with colored marker to ignore accidental carbon marks). We will then circle this scatter shot and estimate the average position of all trials from the given case, and then measure the distance with 1 and 2 meter sticks to determine final horizontal position xx.

After investigating how far the ball travels in the xx direction from a given height in Case 1, we can determine characteristics about the launcher-and-ball system to estimate how far in the xx direction we may expect the ball to travel when launched from a different initial height and/or angle in Cases 2 and 3.

To estimate how far the ball will travel horizontally in Case 2 where we move the launcher to the higher 0° slot (~19 -- 21 cm higher), we can use (5) to determine xcase 2 theoreticalx_{\text{case 2 theoretical}}. However, to do so, we will need to know:

since distance traveled = speed ×\times time.

To determine v0x,case 2v_{0x\text{,case 2}}, we can realize the balls are launched at a zero angle with respect to the floor, similar to that of Case 1. Therefore v0x,case 2=v0x,case 1v_{0x\text{,case 2}} = v_{0x\text{,case 1}}. Thus we can use our data from Case 1 to determine v0x,case 1v_{0x\text{,case 1}}. Since the total velocity for Case 1 is solely in the xx direction, we can also call v0x,case 1v_{0x\text{,case 1}} the launcher’s exit velocity v0,exitv_{\text{0,exit}}. Therefore for today, v0x,case 2=v0x,case 1=v0,exitv_{0x\text{,case 2}} = v_{0x\text{,case 1}} = v_{\text{0,exit}}.

Before we can solve for the exit velocity from Case 1 with (5), we need to know for how long the ball was in the air. If we call the ball’s final height on the floor yy, and we measure the initial height y0y_{0} of the ball, we can use the height difference (Δy=yy0\Delta y = y - y_0) to determine the time tt it took to fall to the floor in the yy direction due solely to gravity gg. Rearranging (7) to solve for tcase 1t_{\text{case 1}}, and knowing that the ball starts at rest in the yy direction (v0y,case 1=0m/s)\text{(}\,v_{0y{\text{,case 1}}} = 0\,\text{m/s}\,\text{)} and treating the floor as ycase 1=0my_{\text{case 1}} = 0\,\text{m}:

y=y0+v0yt12gt2          0=y0+012gt2          2y0,case 1g=tcase 1y = y_{0} + v_{0y}t - \frac{1}{2}gt^{2}~~~~~\Rightarrow~~~~~0 = y_{0} + 0 - \frac{1}{2}gt^{2}~~~~~\Rightarrow~~~~~\sqrt{\frac{2y_{0{\text{,case 1}}}}{g}} = t_{\text{case 1}}

Now that we know the time tcase 1t_{\text{case 1}}, and treating the uncocked position as x0,case 1=0mx_{0{\text{,case 1}}} = 0\,\text{m}, we can use (5) to determine the initial horizontal velocity v0x,case 1v_{0x\text{,case 1}} when released. Reminder, since the total velocity for the zero-angle launch was solely in the xx direction, we are calling this the launcher’s exit velocity v0,exitv_{\text{0,exit}} for the rest of the lab:

x=x0+v0xt          x=0+v0xt          xcase 1tcase 1=v0x,case 1=v0,exitx = x_{0} + v_{0x}t~~~~~\Rightarrow~~~~~x = 0 + v_{0x}t~~~~~\Rightarrow~~~~~\frac{x_{\text{case 1}}}{t_{\text{case 1}}} = v_{0x{\text{,case 1}}} = v_{\text{0,exit}}

Now that we know v0x,case 2=v0,exitv_{0x\text{,case 2}} = v_{\text{0,exit}} at this higher height of Case 2, we focus on tcase 2t_{\text{case 2}}. Similar to before, we can solve for the time by measuring y0,case 2y_{0\text{,case 2}} and plugging it into (10):

2y0,case 2g=tcase 2\sqrt{\frac{2y_{0\text{,case 2}}}{g}} = t_{\text{case 2}}

This leads us back to (5) where we want to solve for xcase 2 theoreticalx_{\text{case 2 theoretical}}. Using the time tcase 2t_{\text{case 2}} and by saying x0,case 2=0mx_{0\text{,case 2}} = 0\,\text{m} in the uncocked position, we find:

x=0+vxt          xcase 2 theoretical=v0,exittcase 2x = 0 + v_{x}t~~~~~\Rightarrow~~~~~x_{\text{case 2 theoretical}} = v_{\text{0,exit}}t_{\text{case 2}}

We will then launch the ball at Case 2’s height and see how accurate we estimated xcase 2 theoreticalx_{\text{case 2 theoretical}}. The experimentally determined xcase 2 experimentalx_{\text{case 2 experimental}} will be measured in the same way as in Case 1 (circling and estimating the center of the scattershot).

○ Angled Trajectory (Background for Case 3)

The third case of this experiment is similar to Case 2 in that we are staying at the higher height, but now investigating angled trajectories and how far in the xx direction do they reach? The large, launcher holder is designed to hold the ball at rest (uncocked) in the same position (both horizontal and height) regardless of angle, so you can treat y0,case 3=y0,case 2y_{0\text{,case 3}} = y_{0\text{,case 2}} as your value for the ball’s initial height and x0,case 3=x0,case 2x_{0\text{,case 3}} = x_{0\text{,case 2}} for the ball’s initial horizontal position (see Figure 1 right).

Example of the upward and downward portions you’ll analyze in Experiment 3 for an angled launch.

Figure 4:Example of the upward and downward portions you’ll analyze in Experiment 3 for an angled launch.

By launching at an upward angle θ\theta (see Figure 4), we are now giving some of the initial velocity to both xx and yy directions. Once again, as there is no acceleration in the xx direction, we will ultimately use (5) to determine xcase 3,theoreticalx_{{\text{case 3,theoretical}}}. However, we need to know both v0x,case 3v_{0x\text{,case 3}} and tcase 3t_{\text{case 3}}. The velocity will merely be the horizontal component of the exit velocity determined in the previous cases:

v0x,case 3=v0,exitcosθv_{0x\text{,case 3}} = v_{\text{0,exit}}\cos{\theta}

To solve for the time, we can characterize motion in the yy direction to determine for how long the ball is in the air. One way to solve for the total travel time is via the quadratic equation with (7) (not shown here). An alternative method to find the total time can be conducted by breaking the trajectory into parts (i.e. tcase 3 = tup + tdownt_{\text{case 3}}~=~t_{\text{up}}~+~t_{\text{down}} as depicted in Figure 4) where tupt_{\text{up}} is the time for upward travel from launch to the peak height, and tdownt_{\text{down}} is the time for downward travel to the floor from that peak height.

UPWARD PORTION tupt_{\text{up}}) Starting with (6), we know the initial velocity in the yy direction is the vertical portion of the exit velocity, v0=v0y,case 3 up=v0,exitsinθv_0 = v_{0y\text{,case 3 up}} = v_{\text{0,exit}}\sin{\theta}. At the peak of the trajectory, the ball goes to rest in the vertical direction such that the final velocity is vy=vy,case 3 up=0m/sv_{y} = v_{y\text{,case 3 up}} = 0\,\text{m/s}, thus:

vy,case 3 up=v0y,case 3 upgtup          0=v0,exitsinθgtupv_{y\text{,case 3 up}} = v_{0y\text{,case 3 up}} - gt_{\text{up}}~~~~~\Rightarrow~~~~~0 = v_{\text{0,exit}}\sin{\theta} - gt_{\text{up}}

leads to

v0,exitsinθg=tup\frac{v_{\text{0,exit}}\sin{\theta}}{g} = t_{\text{up}}

DOWNWARD PORTION tdownt_{\text{down}}) For the downward travel, we can use (7) to determine the time tdownt_{\text{down}} it took to fall from the peak y0,case 3 downy_{\text{0,case 3 down}} to the floor ycase 3 downy_{\text{case 3 down}}. To do this, we need to determine the initial velocity, initial height, and final height.

As mentioned in the upward portion, we know the velocity at the peak of the trajectory in the vertical direction is zero, therefore the downward portion’s initial velocity v0y,case 3 down=vy,case 3 up=0m/sv_{0y\text{,case 3 down}} = v_{y\text{,case 3 up}} = 0\,\text{m/s}. Then, as in previous cases, we also know the final height at the floor is ycase 3 down=0my_{\text{case 3 down}} = 0\,\text{m}.

To determine the initial height for the downward portion, we can realize this equals our final height from the upward portion (i.e. peak height ypeak=y0,case 3 down=ycase 3 upy_{\text{peak}} = y_{\text{0,case 3 down}} = y_{\text{case 3 up}}). We can then use (7) to solve for the final peak height using values regarding the upward portion. We already know our measured initial launch height y0,case 3y_{0\text{,case 3}} as noted from earlier measurements, and the initial velocity in the yy direction v0y,case 3 upv_{0y\text{,case 3 up}} and the time of the upward travel tupt_{\text{up}} as used in (16). Plugging this all in to (7), we find the peak height to be:

ypeak=y0,case 3+v0,exitsin(θ)tup12gtup2y_{\text{peak}} = y_{0\text{,case 3}} + v_{\text{0,exit}}\sin{(\theta)}t_{\text{up}} - \frac{1}{2}gt_{\text{up}}^{2}

Now that we know ypeaky_{\text{peak}}, we can use (7) also to solve for tdownt_{down}. Plugging in variables and substituting the zero floor height and zero initial vertical velocity:

ycase 3 down=ypeak+v0y,case 3 downtdown12gtdown2          0=ypeak+012gtdown2y_{\text{case 3 down}} = y_{\text{peak}} + v_{0y\text{,case 3 down}}t_{\text{down}} - \frac{1}{2}gt_{\text{down}}^{2}~~~~~\Rightarrow~~~~~0 = y_{\text{peak}} + 0 - \frac{1}{2}gt_{\text{down}}^{2}

we subsequently get:

2ypeakg=tdown\sqrt{\frac{2y_{\text{peak}}}{g}} = t_{\text{down}}

HORIZONTAL DISTANCE) Finally, we have the total time tcase 3 = tup + tdownt_{\text{case 3}}~=~t_{\text{up}}~+~t_{\text{down}} from (16) and (19) as well as the initial velocity v0x,case 3v_{0x\text{,case 3}} from (14). Plugging into (5) to determine the theoretical distance xcase 3 theoreticalx_{\text{case 3 theoretical}} the ball with travel in the xx direction for any given non-zero angle θ\theta:

xcase 3 theoretical=v0,exitcos(θ)tcase 3x_{\text{case 3 theoretical}} = v_{\text{0,exit}}\cos{(\theta)}t_{\text{case 3}}

Experimental Procedure

● Procedure Preview

Table 1:Three Primary Experimental Cases with Four Extended Cases

CaseHeightAngle
1Lower00^\circ
2Higher00^\circ
3Higher4545^\circ
AdditionalHigher1515^\circ, 3030^\circ, 6060^\circ, 7575^\circ

● Demo Video: Procedure for Simple Projectle Motion

Overview of Experimental Procedure for the Simple Projectile Motion lab. Notes: All launchers now use the orange balls.

● Case 1 --- Zero-angle Launch

  1. Create a data table for Case 1 including but not limited to:

    • Common data section with the accepted value of gg.

    • Section containing:

      • the ball initial height y0y_{0}

      • the ball height’s estimated uncertainty δy0\delta y_{0}

      • experimental overall distance xx

      • experimental distance estimated uncertainty δx\delta x based on the radius of the circle/ellipse you draw around the scattershot of all launches for case

    • Additional sections for derived time of the trajectory tt and initial velocity v0,exitv_{\text{0,exit}}.

  2. Place the marble launcher in the holder in the lower 0° slot (uncocked to represent where the ball will be once the piston is no longer accelerating the ball up to speed, and the ball is released).

  3. Measure the height the ball will fall (as seen in Figure 2); place the ball into the launcher as the initial height y0y_{0} is measured from the bottom of the ball to the floor (though the bottom of the inside of the barrel can also be used as the bottom of the ball location if that is easier to measure). Ensure the ruler you use has the zero meter end on the floor. Use a plumb bob to find and note the ball’s initial position on the floor.

  4. Conduct a few test launches by pulling the piston back to the denoted notch for your launcher. Take mental note of where the ball is generally landing and ensure it’s generally consistent.

  5. Get some pieces of white printer paper and tape them in the approximate landing zone, and place pieces of carbon paper on top (no need to tape that one since you’ll be moving it between trials) so the ball can mark up the white paper when it lands. Materials available at the table towards the middle of the room. Use this same set of paper throughout all cases.

  6. Conduct 40 -- 45 launches onto the paper/carbon paper. Make sure the launcher plate is the whole way into the slot to ensure a consistent starting position. Between each launch, move aside the carbon paper and mark each dot with a marker or something else that makes it apparent which dots are your data points for this case. Additional markers should be available on the table towards the middle of the room.

  7. Draw a rough circle/ellipse surrounding the scattershot of all launches across all group members for the current case. Flashlights are available at the table towards the middle of the room that could asist in drawing your circle/ellipse. (e.g. Figure 5) Something to consider when determining how large your circle/ellipse should be: remember that we’ve used the standard deviation (the spread of the data relative to the mean) as a stand-in for uncertainty in previous labs. The standard deviation typically includes about two-thirds of the data. Therefore, for use in the following steps, you can estimate your average horizontal distance and uncertainty by drawing a circle/ellipse that encompasses roughly two-thirds of your data, centered around the densest region of your scattershot. Mark the center of your circle/ellipse with a cross hair to represent your average horizontal distance.

  8. Measure the single value for the experimental distance xx from the center of the ball at rest in the barrel (uncocked) to the cross hair center that you drew in your scatter shot on the floor. To translate the initial location of the ball in the barrel to the floor, use a plumb bob to make a straight line down to the floor for x0,x_0\text{,} from which you can more easily measure final distance xx.

  9. From your circle around your scattershot, estimate and measure your uncertainty in horizontal distance δx\delta x, effectively the radius of your circle/ellipse. (if coming from Case 2, return to step 15; if Case 3, return to step 26)

● Case 2 --- Zero-angle Launch at a Higher Height

  1. Create additional data section for Case 2 including but not limited to:

    • Height of the ball at the higher 0° slot height y0,case 2y_{0\text{,case 2}}

    • Time of the trajectory from a higher height tcase 2t_{\text{case 2}}

    • Theoretical distance xcase 2, theoreticalx_{\text{case 2, theoretical}}

    • Experimentally measured distance xcase 2, experimentalx_{\text{case 2, experimental}}

    • Estimated uncertainty in the experimental distance δxcase 2, experimental\delta x_{\text{case 2, experimental}} (essentially ± the radius of the circle drawn around your scattershot)

    • Difference (magnitude) between the theoretical and experimental xx distances

  2. Move the marble launcher to the higher 0° slot and remeasure the initial height y0,case 2y_{0\text{,case 2}} (see Figure 3). Use a plumb bob to ensure you measure vertically.

  3. Now, calculate the theoretical distance xcase 2, theoreticalx_{\text{case 2, theoretical}} using (10) -- (13).

  4. Add paper as needed. Before any launches from the higher height, draw a bullseye at the theoretical distance you expect the balls at the higher height to land to visually see how close we get. You can draw both a cross hair for the distance and estimate how big the scatter will be (to discuss later in Post-Lab Submission --- Interpretation of Results). See example in Figure 3.

  5. Conduct a few test launches by pulling the piston to the same notch you’ve been using in previous case(s) to ensure you have paper coverage for the landing zone. Additionally, by using the same notch, you will also be able to use the same exit velocity as previously determined for the launcher. Add paper as needed.

  6. If need be, tape additional paper in the location from the test launches, while still including paper from previous case(s). Place (no tape needed) a piece of carbon paper on top (no need to tape that one) so the ball can mark up the paper when it lands.

  7. On the same set of white paper on the floor as you retrieved in step 5, repeat steps 6 to 9 to experimentally determine the horizontal distance and its uncertainty when launched from the higher height (i.e. xcase 2, experimentalx_{\text{case 2, experimental}} and δxcase 2, experimental\delta x_{\text{case 2, experimental}}).

  8. Calculate the difference (magnitude, not percent) between your theoretical and experimental values of xcase 2x_{\text{case 2}}.

● Case 3 --- Angled Trajectory at a Higher Height

  1. Create additional data section for Case 3 including but not limited to:

    • Common data section with the accepted value of gg and any values you will need from previous cases to determine the theoretical horizontal distance at a given angled launch ((14) to (20)).

    • Additional sections for:

      • Theoretical distance xcase 3, theoreticalx_{\text{case 3, theoretical}}

      • Experimentally measured distance xcase 3, experimentalx_{\text{case 3, experimental}}

      • Estimated uncertainty in the experimental distance δxcase 3, experimental\delta x_{\text{case 3, experimental}} (essentially ± the radius of the circle drawn around your scattershot)

      • Difference (magnitude) between the theoretical and experimental xx distances

  2. Place the marble launcher in the holder in the 4545^\circ slot (see Figure 3).

  3. Use your previously measured y0,case 2y_{0\text{,case 2}} as the height the ball will fall for any angled launches (e.g. y0,case 2=y0,case 3y_{0\text{,case 2}} = y_{0\text{,case 3}}) as the large holder is designed to release the ball from the same position regardless of angle.

  4. Calculate the theoretical distance xcase 3, theoreticalx_{\text{case 3, theoretical}} using (14) to (20).

  5. Before any launches from the higher height for the non-zero angle, draw a bullseye at the theoretical distance you expect the balls to land to visually see how close we get. You can draw both a cross hair for the distance and estimate how big the scatter will be. Add paper as needed.

  6. Conduct a few test launches by pulling the piston to the same notch you’ve been using in previous case(s) to ensure you have paper coverage for the landing zone. Additionally, by using the same notch, you will also be able to use the same exit velocity as previously determined for the launcher.

  7. If need be, tape additional paper in the location from the test launches, while still including paper from previous case(s). Place (no tape needed) a piece of carbon paper on top (no need to tape that one) so the ball can mark up the paper when it lands.

  8. On the same set of white paper on the floor, repeat steps 6 to 9 to experimentally determine the horizontal distance and its uncertainty when angle-launched from the higher height (i.e. xcase 3, experimentalx_{\text{case 3, experimental}} and δxcase 3, experimental\delta x_{\text{case 3, experimental}}).

  9. Calculate the difference (magnitude, not percent) between your theoretical and experimental values of xcase 3x_{\text{case 3}}.

  10. If there are additional angles assigned, move the marble launcher to the respective angle and repeat steps 6 to 9 as needed, as well as step 27.

  11. TAKE A PHOTO OF ALL YOUR DATA (paper with all your marble impacts measured and labeled across all cases).

Post-Lab Submission --- Interpretation of Results

● Finalized Spreadsheets

● Post-lab Writeup

The Whiteboard

Examples of what experimental data may look like including uncertainty estimations.

Figure 5:Examples of what experimental data may look like including uncertainty estimations.