1) The document describes the components and steps for solving 2-dimensional projectile motion problems. 2) It explains that projectile motion problems can be broken into independent horizontal and vertical components using velocity and acceleration. 3) The document provides the key steps to solve all 2D projectile motion problems: (1) break velocity into x and y components, (2) use displacement to find time, (3) use time to find other displacement, (4) find maximum height, and (5) find final velocity.

1. 2-Dimensional Motion
Horizontal Vertical
What affects
the ball?
Gravity! (No
The only thing to affect friction)
the ball is friction, and we
don’t look at that! g = -9.8m/s2
Velocity is always
constant!

2. Combine Horizontal and Vertical
components are independent!
the two (They don’t affect each
other. It doesn’t matter how
fast the object is going in the
x direction, gravity still acts
evenly on the object.) [Quarters]
So … all these problems are at
least 2 steps: one step in the y
direction, one step in the x
direction.
time is what combines the two.

3. Steps for all these problems
1. Break the velocity into x and y components
vix=vicosθ, viy=visinθ
2. Use the x or y displacement to find time
(x = vit + ½at2)
3. Use the time to find the other displacement:
x or y. (x = vit + ½at2)
4. Find the maximum height. (vf2 = vi2 +2ax)
5. Find the final velocity.

4. Horizontal Problem
A plane flying horizontally (0°) at 91 m/s (204
mph) drops a bag of food supplies from an
altitude of 812 m (0.5 mile). Find:
a. Initial x & y velocities
b. Time required for the food supplies to fall.
c. Range of the projectile (x distance)
d. Final Velocity of the food supplies just
before hitting the ground.

5. Givens: 1. Break velocity into components.
vi = 91 m/s vix = 91 cos 0 = 91
y = -812 m viy = 91 sin 0 = 0
2. Use disp. to find time.
ay = -9.8 m/s2 x = vit + ½at2
-812=0 + ½ (-9.8)t2
t = 12.9 sec
3. Use time to find disp.
x = v ix t
4. Maximum Height x = 91 (12.9) = 1174 m
Easy in this problem!
y = 812 m

6. 5. Find final velocity
vix = vfx remember x velocity stays constant
vfy = viy + at remember y velocity is affected by gravity
vfx= 91m/s
v vf = vi + at
y y
θ = tan −1 fx
vfx
v fy vfy = 0 + -9.8 (12.9)
vfy = -126 2
−1 91 vf = √(vfy2 +vfx )
θ = tan
126 vfy vf vf = 155 m/s
θ = 35.8 vf = 155 m/s at 306°
35.8 + 270 = 306

8. Angle Problem
An arrow is shot from a bow with an initial velocity
of 35 m/s and an initial angle of 30°. If the arrow
lands in a target at the same height it was shot,
find:
a. How long is the arrow in the air?
b. How far would a target be placed in order for the
arrow to hit the bulls eye?
c. the maximum height the arrow travels?
d. the arrow’s final velocity?

9. Givens: 1. Break velocity into components.
vi = 35 m/s vix = 35 cos 30 = 26.0 m/s
y=0m vi = 35 sin 30 = 17.5 m/s
2. Use disp. to find time.
y
ay = -9.8 m/s2 x = vit + ½at2
0 = (17.5)t+ ½ (-9.8)t2
t = 1.89 sec
3. Use time to find disp.
4. Maximum Height x = v ix t
vf2 = vi2 +2ax
0 = 17.52 + 2(-9.8) y x = 26.0 (1.89) = 49.1 m
y = 15.6 m

11. Sample Problem
A disgruntled physics student throws their
book off a 10 m high bridge. If they throw it
at 5.4 m/s at 28°, then find:
a. vix & viy
b. time in the air
c. range (x-disp)
d. maximum height
e. final velocity

12. a. vix = vi cos θ = 5.4 cos 28 = 4.77 m/s
viy = vi sin θ = 5.4 sin 28 = 2.54 m/s
b. y = vit + ½at2 Quadratic
-10 = 2.54 t + ½(-9.8)t2 Equation
0 = -4.9t2 + 2.54t +10 −b ± b 2 − 4ac
2a
−2.54 ± 2.542 − 4(−4.9)(10)
2(−4.9)
c. x = vt
−2.54 ± 14.2
2(−4.9) x = 4.77(1.71)
t = −1.19 or 1.71 x = 8.16 m
(Go with the + time or the greater time!)

13. d. maximum height e. final velocity
vf2 = vi2 + 2ax vfx = vix = 4.77 m/s
0 = 2.542 +2(-9.8)y
vfy = viy + at
y = 0.329 m
= 2.54 + -9.8 (1.71)
vfy = -14.2 m/s
4.77
vf = 4.77 2 + 14.22 = 15.0 m/s
14.2  4.77 
θ =tan-1 ÷ = 18.6° + 270 = 288°
 14.2 