1. Projectile Motion
An object solely under the influence of gravity that possesses a combination of constant horizontal velocity & accelerated vertical velocity, so that it moves along a parabolic trajectory path Ex:
Not:

2. Why are the paths parabolas?
Because in the vertical direction, the object is still in free fall, so there is still a power relationship between the distance it falls and the amount of time passed, which shows up as a parabola on a graph.
Remember Δy α t2 ???
and t Δy
0 0
1 5
2 20
3 45
4 80

3. So then which ball wins on the PM demonstrator – dropped or launched?
The most important thing to remember about projectile motion is that the horizontal motion doesn’t affect (is completely independent from) the vertical motion, and vice versa.
So then which ball wins on the PM demonstrator – dropped or launched?
NEITHER! The fact that the launched one has a horizontal velocity does not affect the rate at which it falls. It falls at the same rate as the one that is simply dropped, so they reach the floor at the same time.
Another way to think of it is that even though the launched ball travels a further (horizontal) distance, it has some extra (horizontal) speed, to get it there, but vertically, both balls are identical.
go back to previous diagram to see…

4. Velocity Vectors and Components for a Projectile
No “i” or “f” subscripts on vh
All instantaneous v’s are tangent to the trajectory path
Symmetry in the instantaneous v’s :
so vi = vf ; θi = θf ; vB = vD ; θB = θD
because vh is constant & vv’s are symmetrical too.
What happens if we cover 1st ½ of the trip, to point C?
Path of a projectile fired with initial velocity v0 at angle  to the horizontal. Path is shown in black, the velocity vectors are green arrows, and velocity components are dashed.

5. Projectile’s Fall from g-Free Path
For any projectile, the vertical distance it falls beneath its imaginary straight line path, if there were no gravity, is the same as the vertical distance it would fall if simply dropped in free fall, from rest, since horizontal and vertical motions are independent from each other. 
Path of a projectile fired with initial velocity v0 at angle  to the horizontal. Path is shown in black, the velocity vectors are green arrows, and velocity components are dashed.

6. Range of a Projectile its horizontal displacement complimentary angles
produce equal ranges:
Time spent in the air by
a projectile depends solely on vv, since
vv alone is affected by gravity, so
vv alone determines how high the projectile will go, which in turns determines how much time it will take to go up & then back down.
If it were up to vh, the object would go forever!
Also, we define an object’s range as to the strike point, so that doesn’t include any rolling the object might do once it hits the ground.
Example 3–8. (a) The range R of a projectile; (b) there are generally two angles 0 that will give the same range. Can you show that if one angle is 01, the other is 02 = 90° - 01?

7. Angle’s Affect on Range of a Projectile
Steeper angles (like 75°) have a large vv so it will go high and be in the air a long time, but it won’t travel very far horizontally since vh is so small for a large angle .
Shallower angles (like 15°) are the opposite – they have a large vh, but a small vv, so it won’t spend much time in the air, and therefore it still can’t go far.
45°, where vh = vv, maximizes the range, since we get a balance between a sizable vh & vv.
Example 3–8. (a) The range R of a projectile; (b) there are generally two angles 0 that will give the same range. Can you show that if one angle is 01, the other is 02 = 90° - 01?

8. How to Solve Projectile Motion Problems
You must keep all vertical and horizontal “given” information separate! Do this by using subscripts of v & h on all velocity variables, all the way through the entire problem. Never put a horizontal variable into a vertical equation, never put a vertical variable into a horizontal equation!
If you need to work with horizontal variables to find something, the only equation you have is vh = Δx/Δt Why??
If you need to work with the vertical variables to find something, you can use any of the 5 constant acceleration equations: ex: vfv = viv + aΔt Why??
Know that the time the projectile spends in the air is a single value, not dependent on either a vertical or a horizontal view – it is the same for both the vertical and horizontal direction.