Consider a ball dropped from a hill of height h.
+y
a = -g
y = 0
What is y0 (initial position) and what is a (acceleration)?
A)  y0 = +h, a=-g
B) y0 = -h, a=+g
C) y0 = 0, a=+g
D) y0 = -h, a=-g
E) y0 = 0, a=-g
L6 M 9/8/14 1 Kinematics in 1D (Ch. 2) -- Applications
(1)  Choose a coordinate system (or reference frame): origin &
positive direction.
(2)  Write down what you KNOW. Take care with signs.
(3)  Identify what you SEEK. (4) Find (or derive) the equation you need, linking what you
KNOW to what you SEEK.
(5) Solve for the unknown algebraically first using symbols:
Check units.
(6) Substitute in numbers (in correct units) to get numerical result.
Does the result make sense? Take care with signs.
L6 M 9/8/14 2 Assignments
For this week:
•  Read Ch. 3 of Wolfson and Prof Dubson’s noted.
•  Do CAPA set #2 by Tues eve at midnight.
•  Do Recitation Homework #2 (found on course web site under
Tutorials, Assignments) by your Wed/Thurs recitation.
L6 M 9/8/14 3 CAPA this week:
#4. A
B
Curve is tighter here at turn A than here at turn B. Therefore, the
magnitude of the curvature is larger and the absolute value of
2 is larger. Thus, the magnitude of x- acceleration is
d 2
x / dt
larger.
L6 M 9/8/14 4 Consider a ball dropped from a hill of height h.
y = 0
a = +g
+y
What is y0 (initial position) and what is a (acceleration)?
A)  y0 = +h, a=-g
B) y0 = -h, a=+g
C) y0 = 0, a=+g
D) y0 = -h, a=-g
E) y0 = 0, a=-g
L6 M 9/8/14 5 Kinematics in 1D (Ch. 2)
a = const
✔ v(t) = v0 + at
1 2
✔ x(t) = x0 + v0t + 2 at
✔ v 2 (x) = v02 + 2a(x − x0 )
1
✔ v = v0 + v = v0 + at
2
2
Equation relates
(v,t)
(x,t)
(v,x)
⎯⎯
→
Note:
When position changes
vertically we typically
replace x with y.
x(t) = x0 + vt
Experimental Fact: For all objects in free fall near the surface of the earth: 

2
direction of a = down
a = +9.8 m / s
L6 M 9/8/14 6 Kinematics in 1D (Ch. 2) -- Applications
Example: A person drops from a 4th story window and falls 15 m to a
waiting net. (1) What is her velocity on impact?
x0 = 15 m
+x
a = -g
x=0
x0 = 15 m
x = 0
a = -g
v0 = 0
v(x=0) = ?
Seek
v2 = v02 + 2a (x - x0)
= 0 – 2g(0 – x0) = 2gx0
v
= ± 2gx0 = ± 2(9.8)(15)
= -17.1 m/s
L6 M 9/8/14 Know
(negative)
7 Kinematics in 1D (Ch. 2) -- Applications
Example: A person drops from a 4th story window and falls 15 m to a
waiting net. (2) When will she impact the net?
x0 = 15 m
One way to find this time is to use
x = 0
the following equation:
Know
a = -g
v0 = 0
v(x=0) = -17.1 m/s
With the information given when
Seek
t(x=0) = ?
will she impact the net?
x0
2x0
1 2
2x0
A)
B)
0 = x0 − gt
⇒ t=±
g
g
2
g
2x0
2g
2x0
2(15)
C)
D)
∴t=
=
= 1.75s
g
x0
g
9.8
L6 M 9/8/14 8 Kinematics in 1D (Ch. 2) -- Applications
A ball is thrown straight up (from ground level) with an initial speed
of v0 . Neglecting air resistance and assuming that positive is
upward, which of the following graphs best represents the ball’s
velocity versus time?
A)
B)
v
C)
v
v0
v0
t
v
t
t
v0
L6 M 9/8/14 9 Kinematics in 1D (Ch. 2) -- Applications
A ball is thrown straight up (from ground level) with an initial
velocity of 22 m/s. Neglecting air resistance, how long will it take for
the object to reach a height of 20 m? 1 2
y = y0 + v0t + at
2
1 2
h = 0 + v0t − gt
2
gt 2 − 2v0t + 2h = 0
L6 M 9/8/14 KNOW:
a = -g
v0 = +22 m/s
y = h = +20 m
y 0 = 0 m
+y
h
a = -g
SEEK: t(y = h)? y = 0
10 Kinematics in 1D (Ch. 2) -- Applications
Therefore, we get the following quadratic equation: 2 - 2v t + 2h = 0
gt
0
The solution to this equation gives the time for a ball thrown in
the air to reach a certain height h. But a quadratic equation
at2+bt+c = 0 has two solutions:
−b ± b 2 − 4ac
t=
2a
where a=g, b=-2v0, c=2h
2v0 1
v0 1
2
t=
±
4v0 − 8gh = ±
v0 2 − 2gh
2g 2g
g g
L6 M 9/8/14 11 Kinematics in 1D (Ch. 2) -- Applications
A quadratic equation for the time t at which a ball thrown in the
air reaches a height h:
v0 1 2
t= ±
vo − 2gh
g g
Why are there two solutions to this problem? A)  One is for the time when the object is on the ground and
the other is when it reaches 20 m in the air.
B)  One is meaningless and can be discarded. The other is the
one of interest.
C)  One is the time the object reaches 20 m on the way up and
the other is the time it reaches 20 m on the way down.
L6 M 9/8/14 12 Kinematics in 1D (Ch. 2) -- Applications
In the previous problem, a quadratic equation is found for the
time at which a ball thrown in the air with initial velocity vo
reaches a height h:
t =
v0 ± 1
v 2 − 2gh
At what time will
g g
0
the ball hit the ground?
A)
v0 g
v0 1
2
2v0
t
=
±
v
0 − 2g(0)
B)
g g
g
v0 v0
2v0
v0 1
= ± = 0 or
C)
+
v0 2 − 2gx
g g
g
g g
v0 1
D)
−
v0 2 − 2gx
g g
L6 M 9/8/14 13 CAPA this week:
+y
a = -g
v0
y=0
KNOW: initial speed of each stone: v0
initial height of each stone: y0 = 0
time difference between the stones: t’
they will cross when they reach the same height: y
SEEK: the time at which they cross (t) First stone:
Second stone:
L6 M 9/8/14 1 2
y = y0 + v0t − gt
2
Flies a time t’
less than the
first stone.
1
y = y0 + v0 (t − t ') − g(t − t ')2
2
14