Download 1.624, 3.7, 23.12 Name: Date: Lesson 5 – 3.5 Acceleration Due To

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Name:___________________
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d  vi t 
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gt v f  vi2  2ad g  9.80m / s 2 1.624, 3.7, 23.12
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Date:___________________
Lesson 5 – 3.5 Acceleration Due To Gravity
Introduction to Free Fall
A free falling object is an object that is falling under the sole influence of gravity. Any object that is
being acted upon only by the force of gravity is said to be in a state of free fall.
There are two important motion characteristics that are true of free-falling objects:
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Free-falling objects do not encounter air resistance.
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All free-falling objects (on Earth) accelerate downwards at a rate of 9.8 m/s/s
What is the acceleration of gravity on
Moon :
Mercury:
Jupiter:
Neptune:
Representing Free Fall by Graphs
A position versus time graph for a free-falling object is shown below.
A velocity versus time graph for a free-falling object is shown above.
Example 1:- An object is dropped from a cliff and hits the ground below 4.0 seconds later.
Neglecting air resistance, find:
(a) the height of the cliff.
(b) the velocity of the object just before hitting the ground.
(c) the average velocity during the fall.
(d) the time taken to attain a velocity of -20 m/s.
(e) how far it travels during the third second.
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t
v f  vi  gt d 
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 vi  v f  t
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d  vi t 
1 2 2
gt v f  vi2  2ad
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g  9.80m / s 2
1.624, 3.7, 23.12
Example 2. A ball is tossed vertically upwards. If it leaves the hand 2.2 m above the ground with a
velocity of 32 m/s, find:
(a) the maximum height reached above the ground.
(b) the time to reach maximum height from the time it is released.
(c) the time to fall back to the ground from the maximum height.
(d) the velocity of the ball as it falls back to earth when it reaches the height from which it
was released.
Example 3. A ball thrown vertically upward returns to the starting point in 4.0 s. Find its initial
velocity.
Example 4. A ball is thrown vertically upwards at 25 m/s.
(a) Calculate the velocity after 4 s.
(b) Calculate the height of the ball at 4 s.
(c) How much time passes before the ball hits the ground?
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v
t
v f  vi  gt d 
1
 vi  v f  t
2
d  vi t 
1 2 2
gt v f  vi2  2ad
2
g  9.80m / s 2
1.624, 3.7, 23.12
Example 5. A pebble is thrown vertically downwards from a bridge with an initial velocity of 10.0
m/s and strikes the water in 2.0 seconds. Find the velocity with which the pebble
strikes the water and find the height of the bridge.
Example 6. On a certain asteroid, a steel ball drops a distance of 0.80 m in 2.00 s from rest.
Assuming uniform acceleration due to gravity on this asteroid, what is the value of the
acceleration due to gravity on the asteroid?
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t
v f  vi  gt d 
1
 vi  v f  t
2
d  vi t 
1 2 2
gt v f  vi2  2ad
2
g  9.80m / s 2
1.624, 3.7, 23.12
Figure 2.14
9. Figure 2.14 is a graph of the velocity of a ball thrown straight up by a strong pitcher, as a
function of time. In the first part of the graph (ending at A), the ball is accelerated to 39.2 m/s in
a time of 0.20 s. After the ball leaves the pitcher's hand, it experiences only the acceleration due
to gravity until it is caught in a glove and brought to rest in the hand of the catcher.
(a) What is the acceleration of the ball while it is being thrown?
(b) What is the acceleration of the ball after it leaves the pitcher's hand? (ABC)
(c) What is the acceleration of the ball while it is being caught? (CD)
(d) What point on the graph (A, B, C or D) corresponds with the instant when the ball is at the
peak of its flight? Explain your answer.
(e) Why is the slope of the graph negative as soon as the ball leaves the pitcher's hand?
(f) Why is the graph labeled velocity rather than speed?
(g) How far up did the ball travel?
(h) How far down did the ball fall?
(i) What is the average velocity of the ball for the whole trip from pitcher's hand to catcher's
hand?
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d  vi t  gt 2 v 2f  vi2  2ad g  9.80m / s 2 1.624, 3.7, 23.12
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Assorted Answers: 54 m, 3.3 s, 3.3 s, -32 m/s, 2.1 s, 82.3 m, -65 m/s, 1.8 s, -25 m, 78 m, -39 m/s, 2.0 s,
 2.0 101 m / s , -14 m/s, 5.1 s, 22 m, 2.0 101 m / s ,  3.0 101 m / s , 40. m, 0.40 m / s 2 , 4.9 m, 0 m/s, 15 m,
2.0 102 m / s 2 , -9.8 m / s 2 ,  2.0 102 m / s 2 , 82.3 m
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