Projectile Motion Lab( phet.colorado.edu, click the Physics link,and then run the Projectile Motion lab )

Please, Write or type your data, calculations, and answers.

Background

Any object moving without a source of propulsion is aprojectile, typically something thrown or fired into the air.Neglecting air resistance (also known as “drag”), a projectile nearthe Earth’s surface will experience a constant accelerationg = 9.81 m/s² downward and no accelerationhorizontally. If we take the horizontal direction to be xand the vertical direction to be y (positive upward) thanthe standard constant acceleration equations will be

x = v_0x t + x₀

v_x = v_0x

y = -½ g t² + v_0y t +y₀

v_y = -g t + v_0y

Notes:

• a_x = 0
• Often x₀= 0 depending on how we choose ourorigin location
• A_y = -g (negative because gravityaccelerates things downward)

Again, these equations assume no air resistance and will onlyapproximate the actual motion of objects flying through the air,often an extremely poor approximation (think frisbees andcurveballs). This week’s lab will not include air resistance, but alater lab might.

Introduction

Navigate to phet.colorado.edu, click on the Physics link,then click and run the Projectile Motion lab. Familiarize yourselfwith the simulation; that is, play around for a while.

We’ll start the lab in the “Intro” section. Note you can controlthe height of the cannon, the cannon’s angle, and the initial speedof the projectile. Further, you can choose among nine differentprojectiles, you can turn air resistance on or off, change the DragCoefficient of objects, and change the target location on theground. Because we have no air resistance, all the projectiles willbehave the same and you can choose whichever you like.

Display options include velocity vectors, acceleration vectors,and a meter that shows Time, Range, and Height when placed alongthe projectile’s path (I will call this the TRH meter). Thesimulation controls include buttons to clear, fire, pause, step,and reset.

Push the reset button, you should have a cannon 10 m high, aimedhorizontally (0°), with an initial speed of 15 m/s. We will startby studying the effect of initial speed(v₀). Leave the cannon height(H) at 10 and the angle q at 0, change theinitial speeds to those indicated and find the Time and Range whenthe projectile hits the ground for each case.

The “range” values are the x-positions of theprojectile, the horizontal distance from the cannon. The “height”values are the y-positions, the vertical distance aboveground level. The cannon height (H) is the projectile’sinitial y position (y₀).

H = 10m q = 0°

Initial Speed (v0)	Time (T)	Range (R)
4 m/s
8 m/s
12 m/s
16 m/s
20 m/s
24 m/s

Use the TRH meter to get the time and range values. We want thevalues for when the projectile hits the ground, make sure theheight says “0 m”, anything else and the projectile wasn’t yet atground level. For 24 m/s you may have to temporarily use the zoomcontrol at the top of the window to see the TRH meter.

Q1) Explain why the times didn’t change eventhough we changed the launch speed. [You can always come back laterand change your answer after learning more later in thelab.]

Q2) How would you describe the relationshipbetween ranges and initial speeds?

Now let’s compare these results to those expected based on theconstant acceleration equations.

Because q = 0°, v_0x =v₀cos 0° = v₀ and v_0y = v₀sin 0° = 0

y-motion: y = -½ g t² +v_0y t + y₀

• Initial height y₀ = H
• If T is the time when the projectile hits the ground,then y = 0 when t = T

Put this all in: 0 = -½ g T² + H

Solve for T: T = sqrt(2 H /g) [square root function]

Q3) Calculate T from this equation,show the work including units in every step.

The time to fall to the ground is determined by the y-motion.With the same starting heights and same initial y velocity, thefalling times are all the same. At least for the case of no airresistance, I wonder if that would still be true with airresistance? A question for another time...

x-motion: x = v_0x t+ x₀

• x₀ = 0
• v_0x = v₀
• x = R when t = T

Put it all in: R = v₀T + 0

Thus: R = v₀T = v₀ sqrt(2H / g)

Since, for this case with q = 0°, the time to reach theground below was the same, the horizontal distance travelled wasproportional to the launch speed. For launch angles other than 0°or if we include air resistance, this will no longer betrue. This is a high school level lab! Thank you for yourhelp. I would appreciate if you include your thought process aswell!

Return the initial speed to 15 m/s (or just reset the wholesimulation). Collect data for the Time and Range that theprojectile hit the ground as you vary the launch angle, fill in thetable below. Again, use the TRH meter and make sure your datapoints are for a height of 0 m.

H = 10m v₀ = 15 m/s

Launch Angle (q)	Time (T)	Range (R)
-45°
-30°
-15°
0°
15°
30°
45°
60°
75°
90°

Q4) Of the launch angles tested, whichgave
(a) the longest time in the air?
(b) the shortest time in the air?
(c) the greatest range?
(d) the least range?

While larger angles gave longer times, that did not alwaysresult in a greater range. That’s because high angles mean asmaller x-component of the velocity. Yes, more time in the air butif the horizontal motion is too slow, the extra time won’t givemuch extra range.

When the cannon height is zero (launched from ground level) onflat terrain with no air resistance, the range of the projectileobeys this formula:

R = v₀² sin 2q /g

Analysis of this formula shows that maximum range occurs forlaunch angle 45° and that complementary angles (like 15° and 75° or50° and 40°) give the same ranges. But that result is only for flatcases with no air resistance.

Q5) Calculate the range predicted by thisformula for v₀ = 15 m/s, q =45°, and g = 9.81 m/s². Explain whythe simulation gave a longer range than thisprediction.