
Unit 1.3 Projectile Motion Lesson
Presentation
•
Physics
•
11th - 12th Grade
•
Hard
Standards-aligned
Robert Johnson
Used 10+ times
FREE Resource
2 Slides • 14 Questions
1
2
Multiple Choice
A projectile is launched at 68.0° above the horizontal and at a speed of 55 m/s. How long will it take the projectile to reach maximum height?
1.8 s
3.2 s
4.5 s
5.2 s
3
Multiple Choice
A projectile is launched at 68.0° above the horizontal and at a speed of 55 m/s. What is the maximum height reached by the projectile?
233 m
158 m
2.6 m
133 m
4
Multiple Choice
A projectile is launched at 68.0° above the horizontal and at a speed of 55 m/s. What is the projectile's x-component of velocity at the maximum height?
39 m/s
55.7 m/s
20.6 m/s
32.3 m/s
5
Multiple Choice
A projectile is launched at 68.0° above the horizontal and at a speed of 55 m/s. What is the horizontal displacement at maximum height?
124 m
116 m
222 m
107 m
6
7
Multiple Choice
If h(t)=−16t2+84t+140 , what is the value used for gravity?
-16
-32
-9.8
-4.9
8
Multiple Choice
If h(t)=−16t2+84t+140 , what is the initial velocity?
84 feet
84 feet per second
140 feet
140 feet per second
9
Multiple Choice
If h(t)=−16t2+84t+140 , what is the initial height?
84 feet
84 feet per second
140 feet
140 feet per second
10
Multiple Choice
If h(t)=−4.9t2+18t+43 , what is the value used for gravity?
-16
-32
-9.8
-4.9
11
Multiple Choice
If h(t)=−4.9t2+18t+43 , what is the initial velocity?
18 meters per second
18 meters
43 meters per second
43 meters
12
Multiple Choice
If h(t)=−4.9t2+18t+43 , what is the initial height?
18 meters per second
18 meters
43 meters per second
43 meters
13
Multiple Choice
Write the function that models the path of an object thrown from a height of 12 feet with an intial velocity of 19 feet/sec.
h(t)=−16t2+12t+19
h(t)=−4.9t2+19t+12
h(t)=−4.9t2+12t+19
h(t)=−16t2+19t+12
14
Multiple Choice
Write the function that models the path of an object thrown from a height of 25 meters with an initial velocity of 40 meters/sec.
h(t)=−16t2+25t+40
h(t)=−4.9t2+25t+40
h(t)=−4.9t2+40t+25
h(t)=−16t2+40t+25
15
Multiple Choice
An object's path is modeled by: h(t)=−16t2+150t+100 HOW LONG until max height?
451.6 seconds
4.7 seconds
100 seconds
10 seconds
16
Multiple Choice
An object's path is modeled by: h(t)=−16t2+150t+100 WHAT is the max height?
451.6 feet
4.7 feet
100 feet
10 feet
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