Name: ______________________________
AP Physics 1, Per. _________
Date: ____________
Unit 2 Homework #9
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.
vi  0
vf 
a  9.8m / s 2
y 
t  3s
y  v0 t  1 2 gt 2
y 
1
2
gt 2
 1 2 (9.8)3 2
 44.1m
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
y 
1
2
gt 2
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
vi  0
v f  25m / s
a  9.8m / s 2
y 
t
v  v0  gt
v  gt
25  9.8t
t  2.55 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):
v  gt
if final velocity is doubled, t will be increased by 2xs
e. Find the time required for it to fall 300 m
vi  0
vf 
a  9.8m / s 2
y  300m
t
y  v0 t  1 2 gt 2
y 
1
2
gt 2
 300  1 2 (9.8)t 2
t  7.82s
v 2  v02  2 gy
f. Find its speed after falling 70 m
v0  0
v 2  2 g y
v
 2(9.8)(70)
a  9.8m / s 2
y  70m
 1372m
v  37.0m / s
t
v  37.0m / s
g. if it falls 2 xs farther, by how much will the speed change
Since
2
v  2 gy
v  2 g y
if y is doubled, v will be increased by 2xs:
v NEW  2 g (2y )  2 2 gy  2v
2. A marble dropped from a bridge strikes the water in
6.0 s. Calculate:
v0  0
v
a  9.8m / s 2
y 
t  6s
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)
v  v0  gt
 0  9.8(6)
 58.8m / s
b. the height of the bridge
v 2  v02  2 gy
v 2  2 gy
(58.8) 2  2(9.8)y
y  176.4m
c. If the bridge were two times higher, by how much would the speed of the marble change when
it hit the water?
Since
v 2  2 gy
v  2 g y
if y is doubled, v will be increased by 2xs:
v NEW  2 g (2y )  2 2 gy  2v
d. If the bridge were two times higher, how much longer would it take for the marble to strike the
water? Need the relationship between y and t without any other variables that change. Then
isolate t since you want to know what happens to t:
y  v0 t  1 2 gt 2
y 
t
1
2
gt 2
2y
g
if y is doubled, t will be increased by 2xs:
t NEW 
2(2y )
2y
 2
 2t
g
g
Free Fall with non-zero vo
3. A ball is thrown downward with an initial speed of 20 m/s on Earth.
What is the:
a. acceleration of the object a = -9.8 m/s2
b. displacement during the first 4 s
v0  20m / s
v
a  9.8m / s 2
y 
v0 = -20m/s
y  v0 t  1 2 gt 2
 20(4)  1 2 (9.8)4 2
 158.4m
t  4s
c. time required to reach a speed of 50 m/s
v0  20m / s
v  50m / s
a  9.8m / s 2
y 
t
v  v0  gt
 50  20  9.8t
t  3.06s
d. time required to fall 300 m (Hint: use g=-10m/s2 and factor the quadratic or use the
quadratic formula)
v0  20m / s
v
y  v0 t  1 2 gt 2
a  10m / s 2
y  300m
 300  20t  1 2 (10)t 2
 300  20t  5t 2
t
t 2  4t  60  0
(t  10)(t  6)  0
Solutions are t = -10s and t = 6s
Since there is no negative time, we choose
t = 6s
e. speed after falling 100 m
v0  20m / s
v 2  v02  2 gy
v
a  9.8m / s
y  100m
 (20) 2  2(9.8)(100)
2
 2360
v  48.6m / s
t
v  48.6m / s
At peak
v= 0
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
v0  15m / s
v
a  9.8m / s 2
y 
y  v0 t  1 2 gt 2
t  1s
 (15)(1)  1 2 (9.8)12
 10.1m
b. time required to reach an upward speed of 3 m/s
v0  15m / s
v0 = 15m/s
v  3m / s
a  9.8m / s 2
y 
t
v  v0  gt
3  15  9.8t
t  1 .2 s
c. time required to reach an downward speed of 5 m/s
v0  15m / s
v  5m / s
a  9.8m / s
y 
2
t
v  v0  gt
 5  15  9.8t
t  2.04s
d. maximum height of the rock?
v0  15m / s
v0
v 2  v02  2 gy
a  9.8m / s 2
y 
0  (15) 2  2(9.8)y
y  11.5m
t
e. what is the acceleration of the rock at its maximum height? 9.8 m/s2 downward
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 1/6th the value on Earth)
Need the relationship between ymax and g without any other variables that change (v0 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
v 2  v02  2 gy
v02  2 gy max
y max 
v02
2g
y MOON
if g were 1/6th, ymax will be increased by
5. A rock is thrown straight up with an initial speed of 22 m/s.
a. How long
v02
6v02


 6y 6xs
1
2g
2( g )
6
At peak
v= 0
will it be in the air before it returns to the
v0  22m / s thrower (hang time)?
v
a  9.8m / s 2
y  0
t
y  v0t  1 2 gt 2
0  22t  1 2 (9.8)12
t  4.5s
b. What is the maximum height of the rock?
v0 = 22m/s
v0  22m / s
v0
a  9.8m / s 2
y 
t
y = 0
v 2  v02  2 gy
0  222  2(9.8)y
y  24.7m
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 1/6th the value on Earth)
Need the relationship between t and g without any other variables that change (v0 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:
0  v0 t  1 2 gt 2
2v
t 0
g
if g were 1/6th,  will be increased by 6xs
t MOON 
2v0
2v0

6
 6t
1
g
g
6
6. A vertical punt has a hang time of 3.8 seconds. What was its initial velocity?
v0 
v
y  v0t  1 2 gt 2
a  9.8m / s 2
y  0
OR
At peak
v= 0
0  v0 (3.8)  1 2 (9.8)3.82
v0  18.6m / s
t  3.8s
v0 
v0
v  v0  gt
a  9.8m / s 2
y 
0  v0  9.8(1.9)
v0
v0  18.6m / s
t  1.9s
7. A student throws a worthless lab partner off a 120 m high bridge with an initial
downward speed of 10 m/s
v0  10m / s
v
a  9.8m / s 2
y  120m
v0 = -10m/s
t
y = -120m
a. How long does it take the deadbeat to hit the ground below? (Hint: can use g=-10m/s2 and
factor the quadratic or use the quadratic formula)
y  v0t  1 2 gt 2
 120  10t  1 2 (10)t 2
 120  10t  5t 2
t 2  2t  24  0
(t  6)(t  4)  0
Solutions are t = -6s and t = 4s
Since there is no negative time, we choose
t = 4s
b. How fast is the classmate going at the moment of impact?
v  v0  gt
 10  9.8(4)
 49.2m / s
y = 0
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
v0 = 0
v0
t = 4s
DROP
v0  0
t = 3s
THROW DOWN
v0 
v
v
a  9.8m / s 2
y 
a  9.8m / s 2
y 
t4
t 3
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.
y  v0t  1 2 gt 2
 0  1 2 (9.8)4 2
 78.4m
y  v0t  1 2 gt 2
 78.4  v0 (3)  1 2 (9.8)32
v0  11.4m
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.)
v0  0
y = -113m
v
a
y  113m
t  18.2
y  v0t  1 2 gt 2
 113  0  1 2 g (18.2 2 )
g  0.68m / s 2
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
v0  12m / s
vf 
a  9.8m / s 2
y  15m
t
y = -15m
y = -15m
v0=12m/s
Thrown Up
a) how long it takes the first rock to hit the ground
v0=-12m/s
v0  12m / s
vf 
a  9.8m / s 2
y  15m
t
Throw Down
y  v0 t  1 2 gt 2
 15  12t  1 2 (9.8)t 2
0  4.9t 2  12t  15
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. v  v  gt
0
 12  9.8(3.36)
 20.9m / s
c) how long it takes the second rock to hit the ground
y  v0 t  1 2 gt 2
 15  12t  1 2 (9.8)t 2
0  4.9t 2  12t  15
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)
11. To determine how high a cliff is, a llama farmer drops a rock, and then 0.800 s later, throws
another rock straight down at a velocity of −10.0 m/s. Both rocks land at the same time. How high
is the cliff?
Drop
Throw Down
(0.8 sec later)
v0 = 0
v0=-10m/s
v0  0
v0  10m / s
v1 
v2 
a  9.8m / s
2
y 
a  9.8m / s 2
y 
t1 
 air
0.80.8 s less than rock 1)
(rockt 2 2ist1in
There are only 2 variables for each rock so no matter what equation you write down, there will be 2
unknowns. However, you have 2 equations (rock 1 and rock 2). With 2 equations, you can solve for 2
unknowns as long as the 2 unknowns are the same variables. So you must find relationships between
the unknown variables of rock 1 and rock 2 (y1=y2=y and t2=t1-0.8). Now we can write down an
equation for y for each rock in terms of the same other unknown variable.
Rock1 (dropped)
Rock2 (thrown)
y  v0 t  1 2 gt12
 0  1 2 (9.8)t12
y  4.9t12
y  v0 t 2  1 2 gt 22
 10t 2  1 2 (9.8)t 22
 10(t1  0.8)  1 2 (9.8)(t1  0.8) 2
 10t1  8  4.9(t12  1.6t1  0.64)
y  4.9t12  2.16t1  4.86
 4.9t12  4.9t12  2.16t1  4.86
t1  2.25s
y  4.9t12  4.9(2.25 2 )  24.8m