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Name: ______________________________ Date: ____________
AP Physics 1, Per. _________ Unit 2 Homework
Free Fall Problems
For each of the following, a complete solution will consist of:
a) a well-labeled diagram of the situation
b) a list of all motion variables with givens, labeled with units and appropriate algebraic signs (+, -)
c) a clear presentation by showing the equation used before producing a numerical answer
1. A body falls freely from rest on Earth.
(I will box the formula before plugging numbers in so that the relationship between the variables
can be easily examined)
a. Find its displacement from t = 0 to t = 3s.
b. if it falls 2 xs longer (for 6 s), how much farther will it fall? Need the relationship between
t and y without any other variables that would be affected by increasing t. Then isolate y
since you want to know what happens to y : The relationship is boxed above
If time of fall is doubled, y will be increased by 4xs
c. Find the time for it to reach a speed of 25 m/s
d. Find the time to reach double the speed (50 m/s) Need relationship between v and t without
any variables that would be affected by changing the speed (see boxed equation above):
if final velocity is doubled, t will be increased by 2xs
e. Find the time required for it to fall 300 m
t s
y
a m s
v
v
f
i
2
m
y gt
y vt gt
2 2
1
2 2
1
2 2
1 0
2 2 y ^1 gt
t
y
a m s
v m s
v
f
i
2
- 8 /
t s
t
v gt
v v gt
0
v gt
t
y m
a m s
v
v
f
i
2
t s
t
y gt
y vt gt
2 2
1
2 2
1
2 2
1 0
f. Find its speed after falling 70 m
g. if it falls 2 xs farther, by how much will the speed change
Since
if y is doubled, v will be increased by 2xs:
2. A marble dropped from a bridge strikes the water in
6.0 s. Calculate:
a. the speed with which it strikes the water (this is
the speed right before it hits, when the marble is still in free fall)
b. the height of the bridge
c. If the bridge were two times higher, by how much would the speed of the marble change when
it hit the water?
Since
if y is doubled, v will be increased by 2xs:
t
y m
a m s
v
v
2
0
v m s
v m s
m
v g y
v v g y
2
2 0
2
v g y
v g y
2
vNEW 2 g ( 2 y ) 2 2 g y 2 v
t s
y
a m s
v
v
2
0
m s
v v gt
0
y m
y
v g y
v v g y
2
2
2 0
2
v g y
v g y
2
vNEW 2 g ( 2 y ) 2 2 g y 2 v
e. speed after falling 100 m
4. A rock is thrown upward with an initial speed of 15 m/s on Earth.
What is the:
a. rock's height after 1 sec
b. time required to reach an upward speed of 3 m/s
c. time required to reach an downward speed of 5 m/s
d. maximum height of the rock?
e. what is the acceleration of the rock at its maximum height? 9.8 m/s
2 downward
t
y m
a m s
v
v m s
2
0
v m s
v m s
v v g y
2
2 0
2
At peak
v = 0
v 0 = 15m/s
t s
y
a m s
v
v m s
2
0
m
y vt gt
2 2
1
2 2
1 0
t
y
a m s
v m s
v m s
2
0
t s
t
v v gt
0
t
y
a m s
v m s
v m s
2
0
t s
t
v v gt
0
t
y
a m s
v
v m s
2
0
y m
y
v v g y
2
2 0
2
f. If the rock were thrown upward on the moon with the same initial speed, how much would the
max height change? (on the moon, the acceleration due to gravity is 1/ th the value on Earth)
Need the relationship between ymax and g without any other variables that change (v 0 is
constant so that can be in the equation). Then isolate ymax since you want to know what
happens to ymax : Since, v = 0 at max height, ymax and g are related by
if g were 1/6th, ymax will be increased by 6xs
5. A rock is thrown straight up with an initial speed of 22 m/s.
a. How long will it be in the air before it returns to the
thrower (hang time)?
b. What is the maximum height of the rock?
c. If the rock were thrown upward on the moon with the same initial speed, how much would the
hang time change? (on the moon, the acceleration due to gravity is 1/ th the value on Earth)
Need the relationship between t and g without any other variables that change (v 0 is constant so
that can be in the equation, y is 0 so that can be in equation). Then isolate t since you want to
know what happens to t :
if g were 1/6th, will be increased by 6xs
g
v y
v g y
v v g y
2 0 max
max
2 0
2 0
2
y g
v
g
v y (^) MOON 6 2
2 0
2 0
At peak
v = 0
v 0 = 22m/s y = 0
t
y
a m s
v
v m s
2
0
t s
t
y vt gt
2 2
1
2 2
1 0
t
y
a m s
v
v m s
2
0
y m
y
v v g y
2
2 0
2
g
v t
vt gt
0
2 2
1 0 2
t g
v
g
v tMOON 6
6
1
0
v 0 = 0
v 0
Drop (^) Throw Down
t = 4s (^) t = 3s
y = -113m
8. When a kid drops a rock off the edge of a cliff, it takes 4.0 s to reach the ground below. When she
throws the rock down, it strikes the ground in 3.0 s. What initial speed did she give the rock? (Hint:
there are two problems based on the two sentences. Use the first to find a quantity that can be used
in the second.)
DROP THROW DOWN
y is the same for the dropped and thrown down rock. Find y for the dropped rock and use as
third variable for the thrown down rock.
9. To determine freefall acceleration on a moon with no atmosphere, you drop your handkerchief off
the roof of a baseball stadium there. The roof is 113 meters tall. The handkerchief reaches the
ground in 18.2 seconds. What is freefall acceleration on this moon? (State the result as a positive
quantity.)
2
0
t
y
a m s
v
v
2
0
t
y
a m s
v
v
m
y vt gt
2 2
1
2 2
1 0
v m
v
y vt gt
0
2 2
1 0
2 2
1 0
0
t
y m
a
v
v
2
2 2
1
2 2
1 0
g m s
g
y vt gt
v 0 =-12m/s
Throw Down
y = -15m
Thrown Up
y = -15m
v 0 =12m/s
10. Two rocks are thrown off the edge of a cliff that is 15.0 m above the ground. The first rock is
thrown upward, at a velocity of +12.0 m/s. The second is thrown downward, at a velocity of
−12.0 m/s. Ignore air resistance. Determine
a) how long it takes the first rock to hit the ground
The roots of the above quadratic equation are t = 3.36s and t = -0.91s
Since t cannot be negative, t = 3.36s
b) at what velocity it hits.
c) how long it takes the second rock to hit the ground
The roots of the above quadratic equation are t = -3.36s and t = 0.91s
Since t cannot be negative, t = 0.91s
d) at what velocity the second rock hits.
-20.9 m/s, the same as the rock thrown up since when the rock thrown up comes back down
to its original height, it is traveling down at 12 m/s (same as the rock thrown down)
t
y m
a m s
v
v m s
f
2
0
t
y m
a m s
v
v m s
f
2
0
2
2 2
1
2 2
1 0
t t
t t
y vt gt
m s
v v gt
0
2
2 2
1
2 2
1 0
t t
t t
y vt gt
