Module 5: Proj Motion

Question 1

Range

Answer 1

Total Horizontal Distance (x length)

Question 2

Height

Answer 2

Total Vertical Distance (x length)

Question 3

Proj. Motion 1, Initial Velocity

Answer 3

Initial Velocity = X Velocity

Question 4

Proj. Motion 1, Final Velocity

Answer 4

X velocity and Final Y velocity (pythag w/angle)

Question 5

Proj. Motion 1, Equations

Answer 5

s = ut (range)
v = at (final vertical velo)
s= 1/2at^2 (time, use vertical displacement)

Question 6

Proj. Motion 2, Initial Velocity

Answer 6

Launched at an angle, use pythag

Question 7

Proj. Motion 2, Vertex/Maximum Height

Answer 7

v = u + at, where v=0 for vert velocity, then do s = ut + 1/2at^2

Question 8

Proj. Motion 2, Total time/Range

Answer 8

Double the time taken to reach vertex, use x displacement to find range, use t to find final velocities of x & y

Question 9

Proj. Motion 3, Introduction

Answer 9

Two part equation, cannot double vertex time to find total time.
1st Equation - Is similar to Proj. Motion 2
2nd Equation - Constant falling object

Question 10

Proj. Motion 3, Time

Answer 10

Use the maximum height, to determine s=1/2at^2 (time), then add times together to determine total time

Question 11

Proj. Motion 3, Range

Answer 11

Use x velocity and time (x velocity is constant)

Question 12

Proj. Motion 3, Height

Answer 12

v = u + at where v=0, find time taken to get to max height and use s=ut+1/2at^2

Question 13

Proj. Motion 3, Range

Answer 13

Use x velocity and time (x velocity is constant)