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Physics 1311
Spring 2010
PS #3 Solutions
Page 1 of 9
Physics 1311
Problem Set #3 Solutions
Problem 1:
Stones are thrown horizontally, with the same initial velocity, from the tops of two
diFerent buildings, A and B. The stone from building A lands
2
times as far from the base of the
building as does the stone from building B. What is the ratio of building A’s height to building
B’s height?
This problem is an exercise in proportional reasoning—using the equations we’ve learned to
solve problems without numbers. The stones are thrown horizontally, which means that their
initial
vertical
velocity is zero. Since there’s no force in the horizontal direction, the stones
move horizontally with the same constant speed
v
. This gives us a relationship between the
horizontal range of the stones and the time,
x
=
vt
.
In the vertical direction, the stones experience the vertical force of gravity. Since the initial
vertical velocity is zero, we get
Δ
p
y
=

mg
Δ
t
=
⇒
v
y
=

gt.
Then, using the position update formula in the vertical direction,
Δ
y
=
v
y,avg
Δ
t
=
⇒
Δ
y
=
v
y
2
Δ
t
=

1
2
gt
2
.
Thus the height of the building and the time it takes the stone to fall are related by
h
=
1
2
2
.
We combine this with the previous relation, to obtain
h
=
1
2
g
±
x
v
²
2
=
⇒
h
=
g
2
v
2
x
2
.
This relationship says that the height of the building is proportional to the
square
of the
stone’s range. (The other factors in this expression are constants; they don’t change from
building A to building B.) So if the stone from building A lands 2 times as far away, this
must mean that building A is
4 times as tall
as building B.
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Spring 2010
PS #3 Solutions
Page 2 of 9
Problem 2:
A projectile is launched with an initial speed of
v
0
at an angle
θ
above the horizontal.
It lands at the same level from which it was launched. What is the magnitude of its average
velocity between launch and landing?
θ
θ
Between launch and landing, there is no
net vertical
displace
ment, so the average of the vertical component of the velocity
is zero. The horizontal velocity is a constant
v
0
cos
θ
through
out the motion, so this is the projectile’s average velocity be
tween launch and landing.
Another way to approach this question is to Fnd the aver
age velocity vector by averaging the initial and Fnal velocity
vectors. The initial velocity vector, in component form, is
±
v
i
=(
v
0
cos
θ
)
ˆx
+(
v
0
sin
θ
)
ˆy
.
The Fnal velocity vector will have a magnitude
v
0
(like the initial speed) but will be directed
at an angle
θ
below the horizontal:
±
v
f
v
0
cos
θ
)

(
v
0
sin
θ
)
.
When we average these two vectors, the vertical components cancel out, and the (equal)
horizontal components average together to produce
v
0
cos
θ
.
Problem 3:
Olympic diver Matthew Mitcham springs upward from a diving board that is
3.30
m
above the water. He enters the water at a
76.0
degree angle with respect to the water surface,
at a speed of
9.20
m/s. (a) Determine the magnitude of his initial velocity. (b) Determine
the direction of his initial velocity, in terms of degrees relative to horizontal. (c) Determine his
maximum height above the water.
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This note was uploaded on 02/21/2010 for the course PHYS 1311 taught by Professor Wiegert during the Spring '10 term at University of Georgia Athens.
 Spring '10
 Wiegert
 Physics

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