San Diego Mesa College		Name_________________________
Physics 100 Lab Report			Date __________Time___________
							Partners ______________________
TITLE: Projectile Motion				______________________________
							______________________________
    							______________________________

Objectives:	To examine the two-dimensional vector properties of a projectile and to determine the height and horizontal position where the projectile will pass through its highest point.

Theory:  	The motion of an object at any instant is measured by its velocity. 

Projectile motion can be described as two independent motions; the horizontal (x) component of the motion and the vertical (y) component of the motion.

When air resistance is ignored the horizontal motion is described by:

			x = vox t

The vertical component is the motion of a free falling body with acceleration due to gravity.

When there is no initial velocity in the y direction:

	vy = g t		y = ½ g t2

When there is initial velocity in the positive or upward direction:

	vy = voy - gt		y = voyt - ½ gt2

Remember, a vector has a direction, so which direction is up and which is down must be identified and labeled with + and – signs in the equations (i.e. g = -9.8 m/s2).

Equipment:	Projectile Launcher
		Nylon Ball (Projectile)
		Plunger
		Plumb bob
		Carbon Paper
		Blank Sheet of Paper
		12” Ruler
		Meter Stick
		2.0m Meter Stick
		Hoop Ring and Stand
		Safety Glasses

PART I:	Finding the initial speed of the projectile.

Setup:		

Procedure:	

1.	Set the launcher so it will fire horizontally (0.0 degrees) at the short or medium setting according to your professor’s instructions.
 2.   Hold the end of the string attached to the plumb bob next to the  on the side of the launcher. Let the plumb bob hang straight down; make sure the string isn’t bending around the clamps or the table.
 3.  Lift the plumb bob so that is just barely above the floor.  Place a piece of masking tape on   the floor and mark on the tape where the point of the plumb bob would touch the tape.  This is    your x = 0.0m mark.
4.  Measure the distance from the floor to the bottom of the circle surrounding the  on the launcher.  This is your value for height (y).  Record it in cm on your table.
5. Put the ball in the launcher and use the plunger to push the ball back to the appropriate setting (1 click only for short, 2 clicks for medium)
6. Make sure nothing is in the way of the launcher.  Have one partner stand about 5 or 6 feet away from the launcher to spot where the ball lands.
7. Pull the long string on top of the launcher at a 90o angle to the launcher to fire the projectile.  If you yank hard or pull at angles other then about 90o (like straight back along the launcher) then the launcher itself may move enough to change your results in an erroneous way.  Please don’t hit your lab partner.
8. Tape the blank piece of paper to the floor about where your ball landed.
9. Place the carbon paper on top of it carbon side down.
10. Fire the projectile about 8 to 10 times.  All of them should land on the paper within about two inches of each other.  If not, find out what might be going wrong, fix it and get 8-10 points that are within 2 inches of each other.

Analysis:	

Measure the distance from the x = 0.0m mark to each of the 8 – 10 marks in your 
cluster.
Take an average of the distances by adding up all the distances and dividing by 
the number of marks.  This is your value for x.

Calculate t, vox from the equations below.  Show all work and use proper units!

For the y-motion, use down as a positive y-direction.

		Y Motion					X Motion

y = ___________	(measured)		x = ___________  (measured)
g = +981  cm/s2	(known)		t = ___________ (calculated) 
voy = 0.0 cm/s		(known)	
t = ___________	(calculated)		vox = _________ = vo

CALCULATION OF t:				CALCULATION OF Vox:

					 

					vo = Initial Speed of Projectile: _____________

PART II:	

Objective:  

To predict where the projectile will have its highest point when launched straight up.

Procedure:	

1. Unclamp the launcher; turn it to fire straight up(90.0 degrees).
2. Firmly re-clamp the launcher.  This doesn’t mean tighten it down until nobody else can turn the
     screw, just firmly.
3. Place the hoop ring and stand on top of the table.
4. Readjust the launcher so that it fires at a 90o angle.
5. Calculate the maximum height.
6. Adjust the center of the hoop so that it is at the same height as the maximum height you 
    calculated.
7. Fire the launcher and hopefully it will stop at the center of hoop or 
    close to the center.

Analysis:	

At the maximum height, the velocity in the vertical direction becomes zero.  We 
can calculate the time up to the maximum height and the height(y) that the 
projectile rises above its release point.

v = v- gt   

where the positive y-direction is up and the acceleration due to gravity is in the 
down or negative direction.

Since the gun now points straight up, v =  v 

At maximum height,  v = 0

Thus, 0 = v - gt

Solving for the time up to maximum height yields:  

(1)	 t = v/g  

Use the above equation to solve for the rise time in the space below.  Show all 
work and  use proper units.

In lecture you learned an expression for the distance traveled under a constant 
acceleration.

 d = vt + at/2

If y is the vertical distance, d, the ball rises from the release point to maximum 
height, and a = -g, we get the following expression:

(2) y = vt - gt/2

If we plug into this equation the expression for t that was derived in equation (1) 
above, we get:

(2) y = v /g – ½(g) (v/g) = v/2g 

for the maximum height the ball will rise above the release point.  

Apply equation (2) to calculate the maximum height that the ball achieves above 
the release point in the space below.    Show all work and use proper units.

	
Did the ball stop close to the center of the ring? 

If the ball did not stop exactly at the center, could the difference in air resistance 
between the ball’s trajectory in parts one and two explain the deviation?  

PART 3.

Procedure:

1.	Place the gun so that it will launch the ball at an angle of 45 degrees.  This angle is halfway between the orientations of the gun in parts 1 and 2.
2.	Calculate the maximum height above the release point the ball will achieve with a launch angle of 45.0 degrees and the corresponding horizontal position of the ball when it is at its maximum height.
3.	Place the ring stand and ring such that the ring’s center is the correct height above the release point and the correct horizontal distance from the release point.
4.	Fire the gun and observe if the ball passes through the center of the ring. 

Analysis:

At the maximum height, the vertical component of the two dimensional velocity is zero.  The time required to rise up to the maximum height was calculated in part 2. 

(1) 	 t = v/g

For a launch angle of 45.0 degrees, it can be shown from right triangle trigonometry that 

(3)         v = v = v/.  

Therefore

(4)        t= v/ (g)

The height, h, that the ball will rise above its release point is:

 y = h = vt – 1/2gt

 Show that  h = v/4g  in the space below by applying (3) and (4) to the above expression.

Calculate the maximum  height above the release point in the space below.

The horizontal distance traveled in this time is:

 x = vt = vt/= (v)/ 2g

Calculate the horizontal displacement in the space below.

Place the ring’s center at a horizontal distance x and a vertical position above the release point, h.

Does the ball pass through the center of the ring? 
If the ball did not pass through the center of the ring, check and correct your calculations and launch the ball again. 

SUMMARY