Special case:
Motion under constant acceleration
Free-fall acceleration
For large variations of the
distance R from the
Earth, the accelerations a1
and a2 are very different
a1
R1
a2
Earth
But if the distance variations are of the order of
meters, the acceleration does not change
significantly.
a ~ const
a
• Objects falling down to Earth
increase their speed 9.8 m/s
every second
a = 9.8 m/s2 g
• The value of g is independent
of the object's mass, shape,
density, etc.
Exercise.
A ball is thrown up. The figure shows the magnitude of the ball's velocities as
it passes through the points A, B, C, D and E. Indicate the direction of the
average acceleration along the different sections of the path.
10 m/s
20 m/s
Adapting our equations to the case of free-fall motion
g= 9.8 m/s2
!
!
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#
y= yMAX
t
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y= 0
!!!
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g= 9.8 m/s2
In particular, when t=t*, the previous three equations take the form,
If we knewt*, we could calculate yMAX
Exercise
Calculate the time at which the ball passes through the
coordinates y = 5 m. Calculate also the corresponding velocity.
Y
i%
y= 5m
y= 0
ii % %t
%&'()*t
t= (+, s -
.()
t= (' s -
/.()
REVIEW
CASE: Constant velocity motion v= vo
CASE: Motion under constant acceleration a= ao
Instantaneous
velocity
time
time
Motion under constant acceleration a
x= x0 + vo t + ½ a t 2
v= vo + a t
x0
a
X
v2= vo2 + 2a (x - x0 )
x
(x - xo)
Application of the above formulas to the particular case of
free fall motion
Y
a
a = - 9.8 m/s2
=-g
Y
y= y0 + vo t - ½ g t 2
v= vo - g t
v2= vo2 - 2g (x - x0 )
g = 9.8 m/s2
9.8 m/s2
The following example is to illustrate that one can
choose the origin of the system of reference at any
point (whichever convenient to solve the problem). The
result will be independent of that particular choice
Example
A ball is dropped from the second floor of a building.
How long will it take for the ball to reach the floor?
Y
Option #1
We decide to place the origin of the
reference at the level of the second floor
At t=0:
yo = 0 and vo = o
So the EQ. of motion take the form
y = y0 + vo t - ½ g t 2
= - ½ g t2
v = vo - g t = - g t
y = - ½ g t2
v= -gt
Option #2
We decide to place the origin of the
reference at the level of the first floor
At t=0:
yo = 7.3m and vo = o
So the EQ. of motion take the form
y = y0 + vo t - ½ g t 2
= 7.3 m- ½ g t 2
v = vo - g t = - g t
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4
5
%
4
5
%
6
7
:
Y
Initial speed of the ball with
respect to the elevator is 20 m/s
y(t)
Elevator ascending at constant
speed of 10 m/s
s(t)
o
Ground
At t=0 (when the ball is thrown up)
the elevator is 28 m above the
ground
x ; <
#
or
or
or
or
:
Position of the ball
At t=0 :
yo = 28 m + 2m = 30 m
vo = 10 m/s
+
Velocity
of the
elevator
20 m/s
Velocity
of the ball with
respect to the
elevator
30 m/s
y(t) =
v(t) =
>
<
v2 = ( 30 m/s)2 - 2 g (y - 30m)
x 7
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5
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@
5A4
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3
5
3A4
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@
4
@
5
3A4
%
@
5A4
%
@
BC
4
@
DC
4
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<
y(t) = 30 + 30 t - (1/2)gt2 Position of the ball
s(t) = so + 10 t
y(t)
Position of the elevator
s(t)
When the ball hits the
floor of the elevator :
s(t) = y(t)
30 + 30 t - (1/2)gt2 = 28 + 10 t
0 = 4.9 t2 - 20 t - 2
t = 4.2 seconds
Alternative method
Since the elevator is moving at constant velocity with respect to the
ground, thePH-221 student can completely ignore the reference
attached to the ground and use instead his/her own reference (the one
attached to the elevator).
Warning: This wouldn't be true if the elevator were accelerating)
z
2 m + 20 (m/s) t - (1/2)gt2
- 2 m - 20 (m/s) t + (1/2)gt2 = 0
which gives t = 4.2 seconds
Same answer, as when
using the previous
method