10. Free fall
Free fall acceleration: g=9.8m/s2
Using general equations:
at 2
x  x0  v0 t 
2
v  v0  at
2ax  x0   v 2  v02
v
x v0  v

t
2
Substitute:
a  g
x y
To derive the following equations:
gt 2
y  y 0  v0 t 
2
v  v0  gt
 2 g  y  y 0   v 2  v02
v
y v0  v

t
2
1
Example: A rocket is fired at a speed of 100 m/s straight up. (Neglect air resistance.)
a) How long does it take to go up to the highest point?
b) What is the maximum height?
c) How long does it take to return back (go up and fall back down)?
d) What is the speed of the rocket just before it falls down?
Given:
v0  100m / s.
a) At the highest point
Solution:
v0

b) The maximum height is
hmax  y  y0  v0t a 
c) When it comes back
gt
g v0 g 
v
 v0t a 

;
2
2
2g
2
a

ta  100m / s 9.8 m s 2  10.2s
v  v0  gt  0  v0  gt  ta  v0 g ;
2
2
0
hmax
2

100m / s 

2  9.8 m s
2
 510m
y  y0
t  0
gt2
gt2
y  y0  v0t 
 0  v0t 

2
2
tc  2v0 g
tc  2ta  20.4s
d) The speed it falls down with is
v  v0  gtc  v0  g 2v0 g   v  v0
2
11. Projectile Motion
The trajectory of an object projected with an initial

velocity v0 at the angle  0 above the horizontal
with negligible air resistance.
v0 x  v0 cos
v0 y  v0 sin 
vx  v0 x  const
a   g  9.8m / s
3
Shoot the monkey (tranquilizer gun)
A zookeeper shoots a tranquilizer dart to a monkey that hangs from a
tree. He aims at the monkey and shoots a dart with an initial speed v0.
The monkey, startled by the gun, lets go immediately.
Will the dart hit the monkey?
A. Only if v0 is large enough.
B. Yes, regardless of the magnitude of vo.
C. No, it misses the monkey.
If there is no gravity, the dart hits the monkey…
If there is gravity, the dart also hits the monkey!
4
Continued
If there is no gravity, the dart hits the monkey…
5
Continued
If there is gravity, the dart also hits the monkey!
Note, that it takes the same amount of time to hit the monkey
as in the no gravity case!
6
Continued
This might be easier to think about…
x  v0t
For the bullet:
1 2
y  gt
2
x  x0
For the monkey:
1 2
y  gt
2
7
Example: A rocket is fired at a speed of 100 m/s at an angle 30° above the horizontal .
(Neglect air resistance.)
a) What are the initial values of the x and y components of the speed?
b) How long is the rocket be in the air?
c) What is the distance between the lunch and landing points (assuming that these
points have the same altitude)?
Given:
y
v0  100m / s
  30
v0 y
Solution:
a)
v0 x  v0 cos  100m / s  cos30  86.6m / s
v0

x
v0 x
v0 y  v0 sin   100m / s sin 30  50.0m / s
b) From the example on slide 2, question (c) follows that
t  2v0 y g ;


t  250m / s  9.8 m s 2  10.2s
c) The horizontal motion is uniform, and the distance is
d  v0 xt  86.6m / s 10.2s   883m
8

v
y
12. Circular motion
Definitions:

r
T – period (time it takes to return to the same point)

f  1 T - frequency
  2πf - angular frequency
For any circular motion there is
radial (centripetal) acceleration:
It is directed along radius and to
the center.

arad
x
d
r
v2

r
For uniform circular motion (v=const):
v
d 2r

 2fr  r
t
T
arad   2r
9
12a. Circular motion (continues)
Optional information (if you are curious):
If circular motion is not uniform (v ≠ const)
then there is tangential acceleration:
The tangential acceleration is perpendicular
to the radial acceleration, and the total
acceleration is equal to:
a tan
v
 lim
t  0 t
2
2
a  a tan
 a rad
10
Example: Period of a satellite motion
g
v2
a
R
ag
R
g  9.8m / s 2
R  6400km
v  gR
v  gR 
v2

g
R
2R
R
T
 2
v
g
9.8m / s 6400 km 1000 m / km   8 10 m / s  8km / s
2

6400km1000m / km
T  2
 2  800s  5000s 
2
9.8m / s
3
5000s
 83min
60 min
11
Example: Two balls attached to a string as shown at 0.20 m and 0.40 m
from the center move in circles at a uniform frequency of 20 rpm.
What are their periods, linear speeds, and radial accelerations?
r1  0.20m
r2  0.40m
f  20rpm
T1, 2  ?
v1, 2  ?
a rad 12
1  1 min  1
1

f   20
 T1  T2   3s


f
 min  60s  3s
2  0.20m

v

 0.42m / s
1


d 2r
3s
v

 
t
T
v  2  0.40m  0.84m / s
2

3s

arad
2

 2 
2
a

 0.20m   0.88m / s
 rad 1 

 3s 
2
 2f  r  
2
 2 

2


a

0
.
40
m

1
.
76
m
/
s


 rad 2  3s 

12