FREEFALL
Lecture Notes
Ex: A ball is thrown directly downward with an initial
velocity of 8 m/s from a height of 30 m.
a) With what velocity does the ball strike the ground?
b) When does it strike the ground?
FREELY FALLING OBJECTS
Free fall Acceleration
If the only force on an object is its weight , the object is
said to be freely falling, regardless of the direction of
motion.
All freely falling objects (heavy or light) have the same
acceleration downward , due to gravity.
t=0
t=1s
v=0
v=2s
v=19.6 m/s
v=3s
v=29.4 m/s
v=9.8 m/s
Ex: An object is released from rest. Find t=0
∆y1
the proportion of its displacements
during the first and second seconds.
t=1s
∆y1
=?
∆y 2
∆y2
g= 9.8 m/s2
t=2s
Ans:
Take t=0 as the starting point:
since v0=0 ⇒ ∆y=1/2 at2:
∆y1=1/2 at2= h ⇒
∆y1+∆y2=1/2a (2t)2=4h; so
∆y2=3h
2
In problem solving we generally take g = 10 m /s for
simplicity.
Downward
Upward
v=0
t=3s
v=0
v=10 m/s
t=2s
v=10 m/s
t=2s
v=20 m/s
t=1s
v=20 m/s
t=3s
v=30 m/s
t=0
v=30 m/s
t=0
t=1s
v=v0+gt
∆y=v0t+1/2gt2
∆t ∆y
v
Sebat Kyrgyz – Turkish High Schools
v0
∆y=(v2-v02)/2g
h
t=1s
4h
Ex: (HW )
Find the proportion of velocities.
v2
=?
v1
∆y
v0=0
h
v1
h
v2
Ex: An object is thrown upwards with an initial
velocity of 20 m/s. Find its velocity
a) One second later
b) Three seconds later
Sol:
[First draw the picture; choose
direcions on the picture. We
generally take direction of v0 as
plus.]
Ex: A stone is dropped is dropped from 100 m above
the ground. Find its velocity and height 3 s later.
v0=0
Note: ∆y does not give
the height, it gives the
displacement
v0=0
{The answer is 3. Now, if we
want, we can speak about the
t=2s
1:3:5:7 proportion of the
displacements of an object for each second when
released from rest}
g= 10 m/s2
Problem solving strategy:
1) First choose (+) and (-) directions. Generally
direction of initial velocity is taken as plus.
2) Choose zero level
3) We can use all the old formulas, only replacing a by g.
t=0
g
+
v0
Vector formula:
v=v0+gt
100 m
v=?
[Now put the values into formula by looking at the
picture to decide about the signs.]
h=?
Scalar equation:
v=(+20) + (-10).1 ⇒ v= +10 m/s
v=(+20) + (-10).3 ⇒ v= -10 m/s
Ex: An object is dropped from the top of a building 20
m high.
a) When does the object strike the ground?
b) What is the final speed?
Ex 3: An object released from rest from high above the
ground reaches the ground with 35 m/s speed. Find
initial height of the object.
Note: If a value is given in the question take it as (+) or
(-) by looking at the picture.
1
FREEFALL
Lecture Notes
Ex: A ball thrown vertically upward is caught by the
thrower after 8 s. Find
a) the initial velocity of the ball
b) the maximum height it reaches
If a value is not given (or it is wanted) in the question
take it as plus; if it is minus the formula will find it as
minus at the end.
Ex: An object is thrown upwards with an initial
velocity of 20 m/s. Find its position
a) One second later
b) Three seconds later {Draw the picture after calculation}
Ex: An object is thrown upwards with v= 25 m/s. Find
its displacement after 5 s.
Ex: Find
h2
=?
h1
Note1: The formula finds the displacement, not the
distance.
Note2: We don’t use two formulas for going up and
down, because the acceleration does not change along
the whole motion.
tf 2
=?
tf 1
Ex: An object is thrown upwards with an initial
velocity of 30 m/s. In what height does its velocity
decrease to 15 m/s?
v=0
2g
hmax
tup = t down
tf2
tf1
Ex: A boy throws his father’s wallet upwards from 5th
floor window which is 21 m above the ground. Find
initial velocity of the wallet if it strikes the ground 3 s
later.
Sebat Kyrgyz – Turkish High Schools
hmax =
v02
h1
v
Ex: {Draw the picture} A balloon is 100 m over the
ground going up with 15 m/s when a passenger
releases a stone. Find the height of the stone (above the
ground) 4s later.
Ex: An object is thrown upwards with an initial
velocity of 30 m/s.
a) In what height does its velocity decrease to zero?
b) What is the maximum height of the object?
Ex: Find a general formula for the maximum height
and time of flight.
{Derive the formula:
Ask: ”Why is the velocity zero at the highest point?;
Answer: Because if it was not zero, the object would
go higher and this point would not be the highest”
Then put vf=0 in the formula ∆y=(v2-v02)/2g}
h2
2v
Ex: A student throws his keys vertically upwards to his
friend in a window 4 m above. The keys are caught 0.4
s later by his friend. Find velocity of the keys:
a) when they are thrown
b) when they are caught
Ex: An object released from rest falls on to the ground
in 4 s. What distance does it travel in the last second of
its flight?
Ex: A falling object travels the last 80 m of its flight in
2s. From what height was it released?
v0
t flight = 2tup
2v
= 0
g
MOTION IN TWO DIMENSIONS
[
Note: Unlike the ones before, these formulas are not
vector formulas. You don’t worry about direction.
y
g
Ex: Three objects are thrown upward with v0=10 m/s,
v0=20 m/s, v0=30 m/s. Draw their maximum heights.
v0
∆r
v
x
v=0
h=45m
v=0
v=0
v0=10m/s h=5m
tup=1s
The vector formulas
v=v0+at and ∆r=v0t=1/2at2 can be written seperately
for x and y components of the vectors.
But according to the figure
ax=0; ay=g; Therefore the formulas become:
Horizontal
∆x=v0xt
vx=v0x
h=20m
v0=20 m/s
tup=2s
v0=30m/s
tup=3s
2
Vertical
∆y=v0yt+1/2 gt2
vy=v0y+gt ]
FREEFALL
Lecture Notes
Ex: Find s=? if the bomb is to fall on to the ship.
Ex: Write the velocity and displacement formulas for
an object thrown with v0 in horizontal direction.
v0
250 m/s
g
2000 m
∆y
vy
vx=v0
0
(ax=0)
Which means:
s=?
vx
∆x
∆x=v0t+1/2axt2
15 m/s
Ex: Find velocity and displacement formulas for an
object thrown at an angle from the horizontal.
v
0
∆y=v0yt+1/2 gt2
0
vy= v0y+gt
vx
v0y
v0
vy
θ
v0
v
∆y
v0x
v0
∆x
v0
vx=v0
vy
vy
v
{Interactive physics demo}
v0=40 m/s
v=?
Ex: Find the displacement of the object in 3 s.
Sebat Kyrgyz – Turkish High Schools
Ex: The object is thrown in horizontal direction with
v0=40 m/s.Find the speed of the object 3 s later.
v0=40 m/s
Velocity:
vx=v0x (ax=0)
vy=v0y+gt
(Vector formulas)
Displacement
∆x=v0x t
∆y=v0yt+1/2gt2
(Vector formulas)
Which means:
v=0
v
v0y
v=vx
v0
v
v0
θ v
0x
θ
v0
v0
Therefore:
v02y
hmax =
2g
t flight = 2tup =
2v0y
g
Ex:
v=?
v0=50 m/s
Ex: By how many meters does the horizontally shot
bullet miss the target 80 m away, if its muzzle velocity
is 400 m/s?
θ=53°
A projectile is fired at an angle 53° above the
horizontal with 50 m/s initial velocity.
a) Find its maximum height
b) Find position and velocity 6 s later.
80m
v0=400 m/s
∆y=?
Ex:
v=?
v0=100 m/s
h=?
θ=37°
640 m
At what height does the projectile hit the wall?
3
FREEFALL
Lecture Notes
Ex:
Ex:
v=50 m/s
10 m/s
15 m/s
h=45 m
v0=?
h=40m
a) What is the initial velocity of the projectile
according to the figure?
b) Find the time of flight
s=?
An object is thrown in horizontal direction with 15 m/s
from a balloon going up with 10 m/s when the balloon
is 40 m above the ground. Find s=?
Ex:
v0=100 m/s
θ=37°
h=?
Sol: {The time formula gives a quadratic equation so
let’s use the timeless formula}
R=?
Find maximum height and maximum range of the
projectile in the figure.
Relative motion
[We will investigate relative motion in two parts:
resultant and relative motion]
Ex:
1. Resultant velocity
VAB is the velocity of A with respect to (as seen from) B.
h=80 m
Sebat Kyrgyz – Turkish High Schools
v0=?
θ=?
R=480 m
Find initial velocity and shooting angle.
Ex: At what angle should we throw an object for
maximum horizontal range?
2v0x v0y
range= ∆xmax =v0x tflight = v0x 2v0y/g =
g
range=2(v0Cosθ) (v0Sinθ)/g
range= 2v02(2Sinθ Cosθ)/g =
2v02
g
Sin 2θ
Therefore range is maximum when Sin2θ=1; θ=45°
Also:
R1=R2 when
θ1+θ2=90°
[Because
Sin60=Sin 120]
Ex: What is the
vBC=4 m/s
velocity of the
boy with respect
to ground?
vBG=?
vBC: velocity of
boy with respect to car
vCG: Velocity of car with respect to ground
Ex: What is the
velocity of the
boy with respect
to ground?
vBG=?
Ans:
vBG= 6 + (-4)=2m/s
Therefore:
v AC = v AB + v BC
v
60°
30°
Ex: Which one will fly higher?
Which one will go farther?
vBC=4 m/s
vCG=6 m/s
(vector sum)
Ex: The boat is
heding directly
vWG=6 m/s
across the river.
What is the
velocity of the
boat as seen by
an observer on the ground?
v
v1
vCG=6 m/s
vBW=8 m/s
[You can think of water as a moving long piece of
paper, and the boat is a bug walking on the paper.The
bug will move on the paper irrespective of the paper’s
motion]
v2
4
FREEFALL
Lecture Notes
s=?
Ex:
vW=3 m/s
d = 14 m
vBW=4 m/s
Ex: Maximum
velocity of a
boat in still
vW=3 m/s
d = 500 m
water is
v=5m/s. This
boat sets off
from one side of a river 500m in width, and travels
directly across the river.Find the passage time for the
river if the speed of water is 4 m/s.
2. Relative motion
Ex: What is the
velocity of the
car with respect
to (as seen by)
the boy?
vboy=3 m/s
vcar=5 m/s
Ex: What is
the velocity of
car A with
respect to car
B?
west
vA=70 km/h
Vector
substruction
east
vB=120 km/h
A
Ex: Two ships are moving on
the sea as in figure. What is
the velocity of small ship
according to the big ship?
Find both magnitude and
sirection of relative velocity.
B
Sebat Kyrgyz – Turkish High Schools
Therefore: vrelative = vobject − vobserver
30 km/h
40 km/h
5