Projectile Motion
Day 1:
Introduction to Projectiles
Definitions: Projectile, Trajectory, Apogee/Zenith, Range, Launch Pair
Go to the Projectile Motion Applet and select Run Now.
Complete the following:
1. Modify the initial speed in the menu to determine the speed required to hit the bull's eye.
2. Put the mouse over the bull's eye and drag it to the intersection the trajectory of the projectile and the horizontal line above the bull's eye.
3. Keeping the speed the same, change the angle to find another angle that also hits the bull's eye in the same spot.
4. Change the initial speed, angle and move the bull's eye to determine a general statement to describe the launch pair (two angles that give the same range for a given initial speed).
Horizontally Launched Projectile
Things to know:
Initial y velocity v_{yo} equals zero
a_{y}=g=9.8 m/s^{2}
a_{x}=0 therefore v_{x }is constant
y displacement is maximum at the beginning
Solving these problems:
1. Find the time
90% of the time you will use information in the y direction (vertical) to find time.
Use the equation: x=v_{o}t + 1/2 at^{2}
When using the y direction v_{yo}=0
and a_{y}=g so the equation is y = 1/2 gt^{2}
When using the x direction (only when you are given the range and v_{xo}) a_{x}=0 so the equation is x=v_{xo}t
2. Find apogee/zenith or range
Use the equations above, to find the information that is not given.
For example, if you use the y direction to find the time, solve for the range with the equation x=v_{xo}t
3. Find the impact velocity
Remember v_{x} is constant.
Find vy at landing using the equation, v_{y} = v_{yo} + gt
Use Pythagorean Theorem to find the velocity
Use inverse tangent to find the impact angle.
Example:
An airplane traveling at 150 m/s at altitude of 3000 m drops a package. How long does it take the package to fall? How far away from its release point does it land? With what velocity does the package hit the ground?
1. Find time
y=1/2gt^{2} => t=24.74 seconds
2. Find range
x=v_{xo}t => x=3712 m
3. Find Final velocity
v_{y} = v_{yo} + gt =>
v_{y} = 9.8(24.74) = 242 m/s
Use Pythagoream Theorem => v=285 m/s
Use inverse tangent Angle = tan^{1} (242/150) = 58.2 degrees
Homework Complete worksheet  Click Here
Day 2:
Review Homework
Complete Horizontal Projectile Practice Problems  Click Here
Review Angle Components Complete Practice Problems  Click Here
Trajectories are parabolic and therefore are symmetric. This means that if the trajectory was folded in half at the apogee the two halves would lie on top of each other. This also provides us some similar values for magntudes of the velocities at a given height along the trajectory. Finally, the angles at a given height on the way up and down are the same, however, the angles on the way down are defined as being negative because the vertical velocity is directed towards the ground.
To solve a full trajectory projectile problem, you follow the following steps:
 Using trigonometric functions for right triangles, find the initial horizontal velocity, v_{xo, }and initial vertical velocity, v_{yo}.
 Using symmetry, set the final horizontal and vertical velocity equal to the initial values, BUT remember that the final vertical velocity is negative.
 Using the initial and final vertical velocities, find the time of flight of the projectile.
 Using the fact that the vertical velocity at the apogee is 0 and the initial vertical velocity, find the maximum height of the projectile.
 Using the fact that horizontal velocity is constant and the time found in step 3, find the range of the projectile.
Example:
A pole vaulter runs along the track at a speed of 8 m/s and suddenly jumps up at an angle of 76 degrees to leap over a hurdle. How long is the person in the air? How high does she go? How far does she travel in the air?
1. Find the horizontal and vertical components of the initial velocity.
V_{ox} = 8 cos76 = 1.94 m/s
v_{oy} = 8 sin76 = 7.76 m/s
3. Find time of flight
v_{y} = v_{oy }+ gt => t = 15.52/9.8 => t = 1.58 seconds
4. Find how high the jumper goes.
v_{y}^{2} = v_{oy}^{2}+ 2gy => y = 3.08 m
5. Find the range of the pole vaulter.
x = v_{ox}t => 3.06 m
Homework  Complete Full Trajectory Problems  Click Here
Day 3:
Review Homework
Discuss Partial and Extended Trajectories
Bull's Eye Lab
Homework  Complete Problems 715 Click Here
Day 4:
Review Homewor
Complete Review Problems  Click Here
Homework  Study for Test
Day 5:
Test
