Free Fall Test Review
Important Formulas that will given on the test:
Vf = Vi + at
Vf2 = Vi2 + 2a Δx
Δx = Vit + ½ a t2
Δx = ½ (Vi + Vf)t
Important questions to be answered by the end of the unit:
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What are the basic motion equations for motion at constant acceleration?
How does one use the constant acceleration equations to solve motion problems?
What criteria are necessary to define “free fall” near the Earth’s surface?
What is the commonly used SI value for the acceleration of gravity near the Earth’s
surface?
How does one use the constant acceleration equations to solve motion problems
involving free fall?
Upon completion of the unit:
1. Compare uniform motion to non-uniform motion
a. What does the motion diagram of an object moving with the same speed vs an
object changing its speed compare
Same speed
a= 0m/s2 -­‐ v= 2m/s a= 0m/s2 v= 2m/s v= 2m/s Changing its speed
-­‐ a= 2m/s2 a= 2m/s2
v= 2m/s v= 4m/s v= 6m/s + + 2. Define acceleration and show acceleration on a:
a. Acceleration definition: Acceleration is the change in velocity in a particular
amount of time. Acceleration is caused by a change in the net force
b. Motion diagram:
-­‐ a= 2m/s2 a= 2m/s2
v= 2m/s v= 4m/s v= 6m/s + c. Velocity-Time graph: http://www.thetrc.org/pda_content/texasphysics/eBookData/BookInd-847.html
d. Other graphs for acceleration
3. Use slope formula to find the average acceleration from a velocity time graph
a. Know that the slope of the velocity time graph yields the acceleration of the
object.
b. Velocity-Time graph: http://www.thetrc.org/pda_content/texasphysics/eBookData/BookInd-847.html
4. Describe the difference between positive and negative acceleration and describe
situations that illustrate each situation using motion diagrams
This graph shows that velocity is increasing in the positive direction
Matches examples #1 and #2
Positive acceleration #1
Real Life example: car moving faster in the positive direction
a= 2m/s2 a= 2m/s2
-­‐ + v= 2m/s v= 4m/s v= 6m/s Positive acceleration #2
+ -­‐ a= 2m/s2 a= 2m/s2
increasing speed
v= 2m/s v= 4m/s v= 6m/s Real life example: elevator moving upward
Positive acceleration #3
Real life example: Car moving in the negative direction and slowing down, due to friction on
the positive direction.
a= 2m/s2 a= 2m/s2
-­‐ + v= -­‐2m/s v= -­‐4m/s v= -­‐6m/s This graph shows that velocity is increasing in the positive
direction
Matches examples #1 and #2
Negative acceleration #1
Real life example: Car moving in the positive direction and slowing down, due to friction on
the negative direction.
-­‐ a= -­‐2m/s2 a= -­‐2m/s2
v= 6m/s v= 4m/s + v= 2m/s + Negative acceleration #2
v= 2m/s v= 6m/s v= 4m/s -­‐ + a= -­‐2m/s2 a= -­‐2m/s2
Real life example: Object thrown in the air, moving in the positive direction and
slowing down, due to gravity force acting in the negative direction.
Negative acceleration #3
-­‐ V = -­‐13m/s v = -­‐16 m/s v= -­‐19 m/s a= -­‐3m/s2 a= -­‐3m/s2
Real life example: Object dropped and travels in the negative direction increasing
speed in the negative direct because the force of gravity is acting in the negative
direction.
6. Define acceleration due to gravity and give examples.
a. Acceleration due to gravity: The acceleration of an object due to the force of
gravity acting on the object in the negative direction at a know acceleration.
For example all objects on earth have ”acceleration due to gravity” (g) of
-9.8m/s2, on the moon, objects on earth have ”acceleration due to gravity” (g)
of -1.62m/s2,
b. What are the conditions for free fall?
Only one force (gravity), no air resistance
c. Does the mass matter? How do you know?
No mass does not matter, class demonstrations
http://study.com/academy/lesson/newtons-laws-and-weight-mass-gravity.html
https://www.youtube.com/watch?v=_mCC-68LyZM
7. What does the motion of an object moving though space look like on a Position vs
Time, Velocity vs Time and Acceleration vs Time graph look like?
See all negative acceleration in vertical plane (Negative accel #2, #3)
Problem Solving
8. Calculate the velocity and displacement of an object undergoing constant
acceleration.
Use a = -9.8 m/s2 or -10m/s2
Vf = Vi + at
Vf2 = Vi2 + 2a Δx
Δx = ½ (Vi + Vf)t
Δx = Vit + ½ a t2
9. State and use the common constant acceleration motion equations to solve onedimension motion problems, including that due to the acceleration due to gravity,
flight time and maximum height of objects thrown upward and falling objects.
a. Solving for velocity- dropped objects
i. How fast will a pebble be traveling 3 seconds after being dropped? (29m/s)
Vf = Vi + at
= 0 +(-9.8) (3s)
= -29m/s
ii. A penny dropped into a wishing well reaches the bottom in 1.50 seconds.
What was the velocity at impact?
Vf = Vi + at
= 0 +(-9.8) (1.5s)
= -14.7m/s
iii. A pitcher threw a baseball straight up at 35.8 meters per second.
What was the ball’s velocity after 2.5 seconds? (Note that, although
the baseball is still climbing, gravity is accelerating it downward.)
Vf = Vi + at
= +35.8 +(-9.8) (2.5s)
= +35.8 -24.5
= 11.3m/s
So at 2.5s it will be on the way up, instead of down
iv. In a bizarre but harmless accident, Superman fell from the top of the
Eiffel Tower. How fast was Superman traveling when he hit the ground
7.8 seconds after falling?
Vf = Vi + at
= 0 +(-9.8) (7.8s)
= -76.4m/s
v. A water balloon was dropped from a high window and struck its target
1.1 seconds later. If the balloon left the person’s hand at –5
meters/sec, what was its velocity on impact?
Vf = Vi + at
= -5 +(-9.8) (1.1s)
= -5-10.78
=-15.78m/s
b. Solving for velocity- thrown object
i. An object is thrown up with an initial velocity of 3m/s
1. What distance did the object travel before it stopped?
Vf = Vi + at
0 = +3 +(-9.8)(t)
-3 = -9.8t
-3/-9.8 = t
0.3s = t
Δx = ½ (Vi + Vf)t
Δx = ½ (+3 + 0)0.3
Δx = 0.45m
2. What was the time to the peak?
Vf = Vi + at
0 = +3 +(-9.8)(t)
-3 = -9.8t
-3/-9.8 = t
0.3s = t
3. What was the total time in the air?
Time to peak = 0.3
Time from peak to ground = 0.3s
0.6s = time total
4. What was the total distance it traveled before getting caught at
the same position it was released.
Distance to peak = 0.45
Distance from peak to ground = 0.45
Total distance = 0.9m
5. What was the speed of the object just before it was caught,
why?
-3m/s, since there is no air resistance, the speed of the
object is the same as when it was thrown up, just in a
different direction.
ii. An object is thrown down with an initial velocity of 3m/s
1. What distance did the object travel, if it traveled for 0.3s?
Δx = Vit + ½ a t2
= (-3)(0.3) + ½ (-9.8) (0.3)2
= (-3)(0.3) + ½ (-9.8) (0.3)2
= (-.0.9)+ ½ (-9.8)(0.09)
=(-0.9) + (-0.441)
= -1.341m
2. What was the speed of the object just before it was caught?
Vf = Vi + at
= -3 + (-9.8) (0.3)
= -3 + (-29.4)
= -32.4m/s
c. Solving for distance
i. A stone tumbles into a mineshaft and strikes bottom after falling for
4.2 seconds. How deep is the mineshaft?
Δx = Vit + ½ a t2
= (0)(4.2) + ½ (-9.8) (4.2)2
= (0) + ½ (-9.8) (17.64)
= -86.436m
ii. A boy threw a small bundle toward his girlfriend on a balcony 10.0
meters above him. The bundle stopped rising in 1.5 seconds. How high did
the bundle travel? Was that high enough for her to catch it?
Vf = Vi + at
0 = Vi + (-9.8)(1.5)
0 = Vi +-14.7
+14.7 = Vi
Δx = Vit + ½ a t2
= +14.7 (1.5) + ½ (-9.8) (1.5)2
= +14.7 (1.5) + (-11.025)
= +22.05-11.025
= +11.025
The bundle would have traveled 11.025m but since his girlfriend was 10m above
him. Therefore, it was caught by her at 10m!
iii. A volleyball serve was in the air for 2.2 seconds before it landed
untouched in the far corner of the opponent’s court. What was the
maximum height of the serve?
The volleyball reaches max height 1.1s after being served
(halfway in its travel).
Since I don’t know the speed it was, when it was served… I’m
going to use the speed at the max height as my initial velocity
and assume the ball starts at the peak and find the distance to
the ground. I’m doing this because the distance from serve to the
max height is the same as the distance from the max height to
the ground.
Δx = Vit + ½ a t2
= 0 (1.1) + ½ (-9.8) (1.1)2
= 0 + ½ (-9.8) (1.21)
= -5.929m
d. Solving for time
i. The climber dropped her compass at the end of her 240-meter climb.
How long did it take to strike bottom?
Δx = Vit + ½ a t2
240 = 0 + ½ (-9.8) (t)2
240 = 0 + -4.9(t)2
240/-4.9 = t2
49.0 = t2
7.0s = t
ii. An object falls a distance of 4om, how long was it in the air?
Δx = Vit + ½ a t2
40 = 0 + ½ (-9.8) (t)2
40 = 0 + -4.9(t)2
40/-4.9 = t2
8.16 = t2
2.86s = t