Projectile Motion

Page
2

Slide 15

Velocity

vx = v0

vy = g t

Θ

Slide 16

Final velocity

vx = v0

vy = g t

Θ

Θ is negative (below the horizontal line)

Slide 17

HORIZONTAL THROW - Summary

h – initial height, v0 – initial horizontal velocity, g = -9.81m/s2

Slide 18

Part 2. Motion of objects projected at an angle

Slide 19

Initial position: x = 0, y = 0

Slide 20

x

y

Motion is accelerated

Acceleration is constant, and downward

a = g = -9.81m/s2

The horizontal (x) component of velocity is constant

The horizontal and vertical motions are independent of each other, but they have a common time

a = g =

- 9.81m/s2

Slide 21

ANALYSIS OF MOTION:

ASSUMPTIONS

x-direction (horizontal): uniform motion

y-direction (vertical): accelerated motion

no air resistance

QUESTIONS

What is the trajectory?

What is the total time of the motion?

What is the horizontal range?

What is the maximum height?

What is the final velocity?

Slide 22

Equations of motion:

Slide 23

Equations of motion:

Slide 24

Trajectory

x = vi t cos Θ

y = vi t sin Θ + ½ g t2

Eliminate time, t

t = x/(vi cos Θ)

y

x

Parabola, open down

y = bx + ax2

Slide 25

Total Time, Δt

final height y = 0, after time interval Δt

0 = vi Δt sin Θ + ½ g (Δt)2

Solve for Δt:

y = vi t sin Θ + ½ g t2

t = 0 Δt

x

Slide 26

Horizontal Range, Δx

final y = 0, time is the total time Δt

x = vi t cos Θ

Δx = vi Δt cos Θ

x

Δx

y

0

sin (2 Θ) = 2 sin Θ cos Θ

Slide 27

Horizontal Range, Δx