Projectile Motion

Section: 1401-A Name: Projectile Motion – Lab Report GOAL: (briefly state what experiment(s) will be performed and with what purpose) PROCEDURE 1: Initial Velocity, Time of Flight, and Range In this procedure you will be simulating an object fired horizontally. The website you will be using is: https://phet.colorado.edu/sims/html/projectile-motion/latest/projectile-motion_en.html 1. Navigate to the website, into the “Intro” tab. 2. Use the following initial conditions for the simulation: · Height = 10 m · Initial Velocity = 20 m/s · Angle = 0 degrees · Air resistance is turned on (checked) · Acceleration Vectors, Components is turned on (checked) · Speed is set to “Slow” 3. Fire the object using the red button at the bottom of the screen 4. Use the blue tool in the toolbox and inspect the time of flight and range of the pumpkin once it has hit the ground

5. Repeat this process to fill the values in Table 1 below Table 1: Angle (degrees) Height (m) Initial Velocity (m/s) Time of Flight (s) Range (m) 0 10 20 1.48s 26.48m 0 10 18 1.48s 24m 0 10 16 1.47s 21.48m 0 10 14 1.47s 18.93m 0 10 12 1.46s 16.34m 0 10 10 1.46s 13.71m 0 10 8 1.46s 11.04m 0 10 6 1.45s 8.33m 0 10 4 1.45s 5.59m 0 10 2 1.45s 2.81m 0 10 0 1.45s 0m Use Equation 3a from the theory section to calculate the time of flight: t = 1.43 s Compare the time of flight values from table 1 (experimental values) with the calculated time of flight (theoretical value) by calculating the percent error .

QUESTIONS: 1. How would the horizontal range change if the initial velocity was doubled? Explain (use the equation from the Theory section, if needed). - Horizontal range equals R = (Vo^2sin2 (theta))/ g. this means R is proportional to Vo^2. With this informatic we can conclude if Vo is 2 then R is 4R. This means if the change of the initial velocity was doubled the horizontal range would become 4 times what it originally was. 2. How would the horizontal range change if the height from the ground was doubled? Explain. - As we increase height from the ground, the time the ball will remain in the air will increase, which will increase the horizontal range. 3. How would the horizontal range change if the mass of the ball was doubled? Explain. - The horiztonal range is independent of the mass of the ball, as can be seen from the equation above. Time of flight is given by the formula: T m= 2usin PROCEDURE 2: Launching at an Angle on a Plane In this procedure, you will launch the object at various angles on a level plane. This means that the object will begin and end its motion at the same height (displacement D y = 0) 1. Navigate to the “Lab” tab at the bottom of the page 2. Use the following initial settings · Angle: 25 degrees · Initial Velocity: 15 m/s · Air resistance is on (checked) · Keep all default settings for cannonball object and gravity

3. Use the tools to measure the range of the cannonball and input the information into “Measured Range” in Table 2. 4. Use Eq. 4 from theory to fill in to calculate the Predicted range for the 25-degree angle 5. Repeat for the angles 35-85 degrees and fill in Table 2 Table 2: Initial Velocity (m/s) Angle , q (degrees) Sin(2 q ) Predicted Range (m) [v 0 2 sin(2 q )/g ] Measured Range (m) % Error 15 25 0.766 17.58 m 13.62 m 29.07 % 15 35 0.939 21.56 m 17.84 m 28.1 % 15 45 1 22.94 m 19.48 m 17.8 % 15 55 0.939 21.56 m 18.44 m 16.92 % 15 65 0.766 17.58 m 15.11 m 16.35 % 15 75 0.5 11.47 m 10.17 m 12.8 % 15 85 0.173 3.98 m 2.48 m 60.5 % Based on the data in Table 2, calculate the average percent error between the predicted (PRange) and the measured range (MRange): 82.95% What angle corresponds to the maximum range? Why is this so? (confirm with the equation) The angle that corresponds to the maximum range is angle 45 because it had the longest distance when the ball was shot out of the cannonball. And once I had measured all the ranges from 25 to 85 it was confirmed that angle 45 had the longest distance. CONCLUSION (Part II)

D = vt D = 7.5m V = 2.5m/s T =? T = sqrt (2h/g) 7.5 m = (2.5 m/s) * t T = 7.5m / 2.5m/s = 3.0s 3.0s = sqrt (2h /9.8m/s^2) 9.0s^2 = 2h /9.8m/s^2 H = 9.0s^2*9.81m/s^2 /2 = 44.15m 4. A ball is dropped from a table 1.2 m high. What is the acceleration of the dropped ball just before it hits the floor? The acceleration of the ball would be g= 9.81 m/s^2. 5. The same ball is now projected horizontally from the same table with unknown initial velocity. What is the magnitude and direction of the acceleration of the ball just before it hits the floor? a) zero * b) g, downward c) g, upward d) cannot be determined because the initial velocity is not specified 6. A projectile is launched with an initial velocity of 20 m/s at 25 degrees above the horizontal. A. - find the x-component of the initial velocity: 20cos25 =18.13m/s - find the y-component of the initial velocity: 20sin25 =8.45m/s

B. Find the horizontal range if this is a level projectile. X velocity * time of flight 18.13 m/s * 1.72s =31.26m C. Find the time of flight 2Vy/g =2*8.45/9.8 =1.72sec 7. A projectile is launched with an initial velocity of 20 m/s at 45 degrees above the horizontal. A. Find the horizontal range if this is a level projectile. Vy = 20sin45 Vx = 20cos45 = 14.14 m/s = 14.14m/s 2Vy/g (time of flight) = 2 * 14.14 / 9.8 = 2.89 sec Range: 14.14 * 2.89 (Vx * time of flight) = 40.86m B. Find the time of flight, compare it to the time of flight in Problem 6C, and explain the difference. In problem 6, the time of flight resulted in 1.72 sec, and problem 7’s resulted in 2.89 sec. Both of these problems have the same initial velocity of 20 m/s, but their angles are different. When the projectile is launched at different angles with the same velocity, the vertical and horizontal components of the velocity vectors are changed. In this case, the time of flight is altered by changing the launch angle while keeping the initial velocity constant. Since problem 7 held a higher launch angle, the time of flight increased. Meanwhile, problem 6 held a shorter launch angle; therefore, the time of flight was less. Once you have completed the assignment, save your report as a PDF and upload using the blackboard submission on the lab cou