1. Two Dimensional Motion and Vectors
Projectile Motion

2. Projectile Motion Use components to simplify calculations
Avoid vector multiplication
Projectile motion – the motion of objects moving in two dimensions under the influence of gravity
Projectile – an object thrown or launched into the air and subject to gravity

3. Projectile Motion
Path of a projectile is a parabola if air resistance is disregarded
Any initial horizontal velocity means the object will have horizontal motion
For our purposes, horizontal velocity remains constant
Air resistance would slow it down

4. Projectile Motion
Projectile motion is free fall with initial horizontal velocity
An object in free fall and a projectile released at the same time, from the same height have the same vertical velocity
Would hit the ground at the same time

5. Projectile Motion Vertical motion of a projectile that falls from rest
vyf = g * Δt
vyf2 = 2 * g * Δy
Δy = ½ * g * (Δt)2
Horizontal Motion of a Projectile
vx = vxi = constant
Δx = vx * Δt
Use the Pythagorean theorem and inverse tangent functions to find velocity at any point on projectile motion

6. Projectile Motion
The Royal Gorge Bridge in Colorado rises 321m above the Arkansas River. Suppose you kick a rock horizontally off the bridge. The magnitude of the rock’s horizontal displacement is 45.0m. Find the speed at which the rock was kicked.

7. Projectile Motion Δy = (-321m) Δx = 45.0m g = (-9.81m/s2) vi = vx = ?y x
Δy = (-321m)
Δx = 45.0m
g = (-9.81m/s2)
vi = vx = ?
vx = Δx / Δt
Δt=?
Δt = √(2 * Δy / g)
= √(2(-321) / (-9.81))
=8.04s
=45.0 / 8.04
=5.56m/s

8. Projectiles Launched at an Angle
Must take the angle into consideration in the calculations
vyf = vi * (sinθ) + g * Δt
vyf2 = vi2 * (sinθ)2 + 2 * g * Δy
Δy = vi * (sinθ) * Δt + ½ * g * (Δt)2
vx = vi * (cosθ) = constant
Δx = vi * (cosθ) * Δt

9. Projectiles Launched at an Angle
A zookeeper finds an escaped monkey on a pole. While aiming her tranquilizer gun at the monkey, she kneels 10.0m from the pole, which is 5.00m high. The tip of her gun is 1.00m above the ground. At the moment the zookeeper shoots, the monkey drops a banana. The dart travels at 50.0m/s. Will the dart hit the monkey, the banana, or neither one?

10. Projectiles Launched at an Angle
Δx=10.0m
g=-9.81m/s2
vi=50.0m/s
Δy=4.00m
θ=?
θ=tan-1(opp / adj)
θ=tan-1(Δy / Δx)
θ=tan-1(4.00 / 10.0)
θ=21.8°
Δy=1/2 * g * Δt2
Δt=?
Δt=Δx / (vi * cosθ)
Δt=10.0 / (50.0cos 21.8)
Δt=.215s y
10.0m x
1.00m
4.00m ?

11. Projectiles Launched at an Angle
Δyb = ½ * g * Δt2
= ½(-9.81)(.215) 2
= -.227m
Δyd = vi * (sinθ) * Δt + ½ * g * Δt2
= 50.0(sin 21.8)(.215)+ ½(-9.81)(.215) 2
= 3.76m
Banana fell .227m
Dart rises 3.76m
Banana is now { = 4.77m} above the ground
Dart is now { = 4.76m} above the ground
Answer: Dart hits banana