simple problems, but remember that this is an approximation
and not an exact value.
•
Since the direction of the acceleration is down (y dir), a
y
= g
Kinematic Equations For
Free Fall
Same exact set of equations, but with the following substitutions:
x
→
y and a
y
→
g
y

y
0
=
v
0
y
(
t

t
0
)

1
2
g
(
t

t
0
)
2
v
y
=
v
0
y

g
(
t

t
0
)
y

y
0
=
1
2
(
v
y
+
v
0
y
)(
t

t
0
)
v
2
y
=
v
2
0
y

2
g
(
y

y
0
)
Clicker QuestionA tennis player tosses a ball straight up before hitting the ball to serve. While the ball is in free fall, its acceleration does which of the following?
a)increasesb)increases then decreasesc)decreases then increases d)remains constante)decreases
t (s)ay(m/s2)g+ yy = 0ymax0tfAt t = 0, ball is thrown straight up with initial velocity v0yand is caught at the starting position (y = 0) at t = tfIt has constant acceleration ay= g throughout the entire motionay=gAcceleration vs. Time
t (s)vy(m/s)+ yy = 0ymaxv0yv0y0tf0.5tfOn the way up, the ball’s velocity is positiveAt the highest point, the ball’s velocity goes to zeroOn the way down, the ball’s velocity is negativeThe vt graph is a straight line with slope g and yintercept v0yvy= v0ygtVelocity vs. Time
+ yy = 0ymaxy (m)0ymax0.5tftft (s)The yt graph is a parabola with negative curvature(denoting constant, negative acceleration)The slope is of the yt graph positive for the firsthalf of the motion (going up) and negative for the second half of the motion (going down)y = v0yt12gt2Position vs. Time
Recap
•The acceleration is constant, so there is no need to break the motion into the upward and downward segments when using kinematic equations (ay= g, at all times)•The ball momentarily comes to a stop at its highest point (vy= 0 at the highest point)•Just before the ball is caught, it has the same speed as it was initially thrown up with, but it is now moving in the y direction (i.e., vy= v0y)•The ball spends the same amount of time on the way up as it spends on the way down
Problem Solving Strategy for Free Fall•Draw a picture or diagram of the motion of the object•Define the positive and negative directions and the origin (where on your diagram is y=0? label it!)•Choose an instant to call initial time (t0) and the final time (t)•For example, “t0 is the moment the ball is thrown & t is the moment the ball lands on the ground”•Define all other known and unknown quantities involved •Choose the kinematic equation(s) to use based on what variables are given and what variable you need to solve for. •Write down the equation(s) before plugging in numbers!•Solve the system of equations (# of equations = # of unknowns)
Dealing With Multiple Solutions
•
You may find situations where solving a problem requires
solving a quadratic equation using the quadratic formula.
•
You’ll have to decide which of the two possible solutions
corresponds to the actual solution to the problem.
•
Example: a ball is launched straight up from ground level at
500 m/s. At what time is the rocket’s height 1000m?
y

y
0
=
v
0
y
(
t

t
0
)

1
2
g
(
t

t
0
)
2
1000
m
= (500
m/s
)
t

1
2
(10
m/s
2
)
t
2
(5
m/s
2
)
t
2

(500
m/s
)
t
+ 1000
m
= 0
t
=
500
m/s
±
p
(500
m/s
)
2

4(5
m/s
2
)(1000
m
)
2(5
m/s
2
)
= 2.0 s, 98 s
Dealing With Multiple Solutions