Group Phys 110 Lab 4 Projectile Motionthe Big Ideatesting ✓ Solved

GROUP # _______ PHYS 110 Lab #4: Projectile Motion The big idea: Testing our model for motion Reminders The muzzle velocity refers to the velocity the projectile has as it leaves the launcher. The range refers to the horizontal displacement of the projectile. For 2D motion, the general equations for displacement are: Equation (1) for horizontal motion: Equation (2) for vertical motion: When launched at an angle, the velocity vector is the hypotenuse of a triangle and the sides of the triangle are the components of the velocity. Equation (3) Horizontal component of muzzle velocity: Equation (4) Vertical component of muzzle velocity: For a horizontal launch, the launch angle is zero, which means and .

Accident Investigation – Creating the Model In the mock-up above, which variable is related to the horizontal displacement of the car/projectile? Does the car/projectile leave the bridge at an angle? ( Yes/No ) If yes, what is the launch angle? At the moment of launch -the instant when the car leaves the bridge: is the horizontal component of the car’s velocity (vox) to the right , to the left or zero ? After launch while the projectile flies through the air, is the horizontal acceleration of the projectile to the right , to the left or zero ? You are part of a car accident investigative team, looking into a case where a car drove off a bridge.

You are using the lab projectile launcher to simulate the accident and to test your mathematical model (an equation that applies to the situation) before you apply the model to the accident data. We are assuming we can treat the car as a projectile. Use Equation (1) and your answers to the questions above to get an equation for the horizontal part of the motion. There should be no numbers in your final answer (except any that are part of the equation), but you may substitute any quantities that are zero to simplify your expression. (Hints: In Equation 1, what variable represents the horizontal displacement? Do you need Equation 3?) In the mock-up above, which variable is related to the vertical displacement of the car/projectile?

At the moment of launch -the instant when the car leaves the bridge: is the vertical component of the car’s velocity (voy) up , down , or zero ? After launch while the projectile flies through the air, is the vertical acceleration of the projectile to the up , to the down or zero ? In lab, g refers to the magnitude of the acceleration due to gravity. Which of the following ( A or B ) correctly represents the acceleration due to gravity (magnitude and direction)? (A) (B) Use Equation (2) and your answers to the questions above to get an equation for the vertical part of the motion. There should be no numbers in your final answer (except any that are part of the equation), but you may substitute any quantities that are zero to simplify your expression. (Hints: In Equation 2, what variable represents the vertical displacement?

Do you need Equation 4?) Because we have no time information about the accident, you need to eliminate the variable t from your equations. (Hint: combine your answers from the previous two questions.) Re-arrange your answer to solve for v0 , the speed of the car as it leaves the bridge. (“Solving for v0 †means that the velocity variable v0 is by itself on the left side of the equal sign and all the other variables are on the right side of the equal sign.) Your final answer here is your mathematical model for the car going off the bridge. Motorcycle Stunt – Creating the Model In the sketch above, which variable is related to the horizontal displacement of the motorcycle? Does the motorcycle leave the ramp at an angle? ( Yes/No ) At the moment of launch -the instant when the motorcycle leaves the ramp: is the horizontal component of the motorcycle’s velocity (vox) to the right , to the left or zero ?

After launch while the motorcycle flies through the air, is the horizontal acceleration of the motorcycle to the right , to the left or zero ? Here you are designing a motorcycle stunt where the motorcycle leaves a ramp at an angle θ and lands on another ramp of the same height. You need to know where to place the second ramp. You will make a mathematical model to predict the range (i.e., to predict how far the motorcycle will land from the first ramp). Use Equation (1) and your answers to the questions above to get an equation for the horizontal part of the motion.

There should be no numbers in your final answer (except any that are part of the equation), but you may substitute any quantities that are zero to simplify your expression. (Hints: In Equation 1, what variable represents the horizontal displacement? Do you need Equation 3?) In the sketch above, is the vertical displacement of the motorcycle up , down , or zero ? Does the motorcycle leave the ramp at an angle? ( Yes/No ) At the moment of launch -the instant when the motorcycle leaves the ramp: is the vertical component of the motorcycle’s velocity (voy) up , down , or zero ? After launch while the motorcycle flies through the air, is the vertical acceleration of the motorcycle to the up , to the down or zero ?

Use Equation (2) and your answers to the questions above to get an equation for the vertical part of the motion. There should be no numbers in your final answer (except any that are part of the equation), but you may substitute any quantities that are zero to simplify your expression. (Hints: In Equation 2, what variable represents the vertical displacement? Do you need Equation 4?) Because we have no time information for the motorcycle, you need to eliminate the variable t from your equations. (Hint: combine your answers from the previous two questions.) Re-arrange your answer to solve for R , the range for the motorcycle. Your final answer here is your mathematical model for the motorcycle’s range.

Accident Investigation – Testing your mathematical model You would now use the launcher to shoot a projectile to simulate the car going off the bridge. Use the videos and pictures to do the following: Measure the “height†of your bridge. The projectile will land on carbon paper, creating a mark on the exact landing point and allowing you to measure the range (the horizontal displacement) of your projectile. Record your data in Table 1. (Use averages) Use your data and your mathematical model to calculate the muzzle velocity. Show your calculations in the box below the table.

Table 1: Data for Mathematical Model Range Setting vo (m/s) (measured by Capstone) h (m) R (m) vo (m/s) (calculated with your model) Short Medium Long vo (m/s) (show calculations with your model) Calculate the Percent Difference between the muzzle velocity as measured by Capstone and the muzzle velocity you calculated with your mathematical model: Range Setting Show your work here: PD Short Medium Long QUESTION 1: Is your mathematical model a good predictor of the projectile’s velocity as it leaves the bridge? If yes, what evidence can you provide to support your answer? If not, explain why not and what modifications you need to make to your model. Motorcycle Stunt – Testing the Model PREDICTION 1: Use your mathematical model to predict the range for a projectile launched at 30 degrees using the short range muzzle velocity measured by Capstone in Part 3.

Show your calculation in the calculation box below. Mathematical model: Predicted Range for Projectile EQUATION(s): CALCULATION: You would use the launcher to shoot a projectile at your angle to test your prediction. (Note: A labjack fully extended is about the same height as the projectile launcher.) The photogate measures the muzzle speed. Take your data from the videos and pictures. Measured Range: QUESTION 2: Is your mathematical model a good predictor for the motorcycle’s range? If yes, what evidence can you provide to support your answer?

If not, explain why not and what modifications you need to make to your model. Accident Investigation – Conclusion You’ll now apply the horizontal launch to a car accident example. In this scenario, a car driver named Joshua Horton says he was driving at the speed limit of 45 mph when he hit an icy patch and drove through a guard rail of the 15 m tall bridge, before landing 24 m from the base of the bridge. Testing shows that driving through the guard rail lowers a car’s speed by 10 m/s. QUESTION 3: Given the distance that the car landed from the bridge, determine what speed Joshua was actually driving before leaving the bridge.

Was he telling the truth or lying? Show your calculations here. Motorcycle Stunt – Applying the Model If you are happy with your mathematical model, apply this model to predict the range of the motorcycle. You can assume the ramp will have the same angle used in your prediction. QUESTION 5: A motorcycle stunt driver does a canyon jump stunt.

If the canyon is 100 m wide, will a launch speed of 75 mph at 45 degrees be fast enough to get the driver across the canyon? Justify your answer with a calculation. Lab #4: Projectile Motion Group Worksheet | Page of Lab #4: Projectile Motion Group Worksheet | Page of Lab #4: Projectile Motion Group Worksheet | Page of

Paper for above instructions

PHYS 110 Lab #4: Projectile Motion

Introduction

Projectile motion is a fundamental concept in physics, describing the motion of an object that is influenced solely by the forces of gravity and its initial velocity. This lab focuses on deriving mathematical models to predict motion behavior when a car accident occurs after driving off a bridge and to design a motorcycle stunt jump.

Accident Investigation – Mathematical Model Creation

In our investigation regarding a car accident where a car drove off a bridge, we focused on two major aspects: horizontal and vertical motions.

Horizontal Motion

For horizontal motion, we can denote the horizontal displacement \(R\) as:
\[
R = v_{ox} \cdot t
\]
Where:
- \(R\) is the horizontal range (displacement),
- \(v_{ox}\) is the horizontal component of the car's launch velocity, and
- \(t\) is the time of flight.
The car leaves the bridge horizontally, therefore \(v_{ox}\) is the initial velocity of the car at the moment of launch. Since it is moving to the right, we assume \(v_{ox} > 0\) and there is no horizontal acceleration:
\[
a_x = 0
\]

Vertical Motion

For vertical motion, we use the equation:
\[
h = v_{oy} \cdot t + \frac{1}{2} (-g) \cdot t^2
\]
In this case, \(h\) represents the height of the bridge (15 m). Since the car is launched horizontally, the vertical component of the velocity \(v_{oy} = 0\) at the moment of leaving the bridge, simplifying our equation to:
\[
h = \frac{1}{2} (-g) \cdot t^2
\]
Using \(g = 9.81 \, \text{m/s}^2\):
\[
h = -\frac{1}{2} \cdot 9.81 \cdot t^2
\]

Eliminating Time

To eliminate \(t\), we need to express it from the horizontal motion equation:
\[
t = \frac{R}{v_{ox}}
\]
Substituting \(t\) in the vertical motion equation yields:
\[
h = -\frac{1}{2} \cdot 9.81 \left(\frac{R}{v_{ox}}\right)^2
\]
Rearranging to solve for \(v_{ox}\):
\[
v_{ox} = \sqrt{\frac{9.81 \cdot R^2}{-2h}} \quad \text{(We omit the negative sign since distance cannot be negative)}
\]
This equation serves as our mathematical model for the situation.

Motorcycle Stunt – Mathematical Model Creation

For designing a motorcycle stunt to clear a canyon, we also derive models for horizontal and vertical motions.

Horizontal Motion

The motorcycle leaves the ramp at an angle \(\theta\). The horizontal displacement \(R\) can be expressed as:
\[
R = v_{ox} \cdot t
\]
Where \(v_{ox} = v_0 \cdot \cos(\theta)\).

Vertical Motion

The vertical position is determined by:
\[
h = v_{oy} \cdot t + \frac{1}{2} (-g) \cdot t^2
\]
Here, \(v_{oy} = v_0 \cdot \sin(\theta)\). Thus, substituting gives:
\[
h = (v_0 \cdot \sin(\theta)) \cdot t - \frac{1}{2} \cdot g \cdot t^2
\]

Eliminating Time

From the horizontal motion equation:
\[
t = \frac{R}{v_ox} = \frac{R}{v_0 \cdot \cos(\theta)}
\]
Substituting for \(t\) in the vertical motion equation:
\[
h = (v_0 \cdot \sin(\theta)) \cdot \left(\frac{R}{v_0 \cdot \cos(\theta)}\right) - \frac{1}{2} \cdot g \cdot \left(\frac{R}{v_0 \cdot \cos(\theta)}\right)^2
\]
Simplifying this equation leads to a relation for the range:
\[
R = \frac{v_0^2 \cdot \sin(2\theta)}{g}
\]
This equation predicts the horizontal distance the motorcycle will travel.

Testing the Mathematical Models

Accident Investigation – Prediction of Muzzle Velocity

Using the derived formulas, we substitute the height (\(h = 15 \, \text{m}\)) and distance (\(R = 24 \, \text{m}\)):
\[
v_{ox} = \sqrt{\frac{9.81 \cdot 24^2}{-2 \cdot 15}}
\]
Calculating this provides us the expected velocity of the car when leaving the bridge. If the calculated speed approximately matches observed values (e.g. using Capstone measurement), our model serves as a good predictor.

Motorcycle Stunt – Predicting the Range

Using the launch speed of the motorcycle and the derived equation \(R = \frac{v_0^2 \cdot \sin(2\theta)}{g}\) with a launch angle of 30 degrees:
\[
R = \frac{(v_0^2) \cdot \sin(60)}{9.81}
\]
Further calculations on the ranges will allow for comparisons with experimental data to identify the model’s accuracy.

Conclusion

Through this lab, we modeled projectile motion for both a car accident scenario and a motorcycle stunt jump, derivative equations predict ranges and speeds. The application of these formulas tests the validity of our mathematical model, revealing its accuracy based on empirical data gathered during experiments. However, discrepancies observed can warrant refinement in model parameters or conditions.

References

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5. Hibbler, R.C. (2014). Engineering Mechanics: Dynamics. Pearson.
6. Tipler, P.A., & Mosca, G. (2014). Physics for Scientists and Engineers. W. H. Freeman and Company.
7. Wolfson, R. (2016). Physics. Addison-Wesley.
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