motion. The figure shows the launch
conditions
and
an
appropriate
coordinate system. The speeds
v,
v
x
,
and
v
y
are related by the Pythagorean
Theorem.
0
v
r
θ
0
h
y
x
v
r
j
v
y
ˆ
i
v
x
ˆ
The squares of the vertical and
horizontal components of the
object’s velocity are:
gh
v
v
y
2
sin
0
2
2
0
2
−
=
θ
and
0
2
2
0
2
cos
θ
v
v
x
=
The relationship between these
variables is:
2
2
2
y
x
v
v
v
+
=
Substitute for
v
x
and
v
y
and
simplify to obtain:
(
)
gh
v
gh
v
gh
v
v
v
2
2
sin
cos
2
sin
cos
2
0
0
2
0
2
2
0
0
2
2
0
0
2
2
0
2
−
=
−
+
=
−
+
=
θ
θ
θ
θ
79
••
At
2
1
of its maximum height, the speed of a projectile is
4
3
of its initial
speed. What was its launch angle? (Ignore any effects due to air resistance.)

Chapter 3
228
Picture
the
Problem
In the absence of
air resistance, the projectile experiences
constant acceleration during its flight
and we can use constant-acceleration
equations to relate the speeds at half the
maximum height and at the maximum
height to the launch angle
θ
0
of the
projectile.
θ
0
v
r
0
y
x
h
2
1
h
The angle the initial velocity makes
with the horizontal is related to the
initial velocity components.
x
y
v
v
0
0
0
tan
=
θ
(1)
Substitute
∆
y
=
h
and
v
y
= 0 in the
equation
y
a
v
v
y
y
Δ
+
=
2
2
0
2
to obtain:
2
0
2
0
gh
v
h
y
y
−
=
⇒
=
Δ
(2)
Write the equation
y
a
v
v
y
y
Δ
+
=
2
2
0
2
for
h
y
2
1
Δ
=
:
2
2
2
2
0
2
h
g
v
v
h
y
y
y
−
=
⇒
=
Δ
(3)
We are given that
0
4
3
v
v
=
when
h
y
2
1
Δ
=
. Square both sides and
express this using the components of
the velocity.
The
x
component of the
velocity remains constant.
(
)
4
3
2
0
2
0
2
2
2
0
y
x
y
x
v
v
v
v
+
⎟
⎠
⎞
⎜
⎝
⎛
=
+
(4)
where we have used
v
x
=
v
0
x
.
(Equations 2, 3, and 4 constitute three equations in the four unknowns
v
0x
,
v
0y
,
v
y
,
and
h
.
To solve for any of these unknowns, we first need a fourth equation.
However, to solve for the ratio (
v
0y
/
v
0x
) of two of the unknowns, the three
equations are sufficient.
That is because dividing both sides of each equation by
v
0x
2
gives three equations and three unknowns
v
y
/
v
0x
,
v
0y
/
v
0x
, and
h
/
.
2
x
0
v
Solve equation (3) for
gh
and
substitute in equation (2):
(
)
2
2
0
2
0
2
h
y
y
v
v
v
−
=
⇒
2
2
0
2
y
y
v
v
=
Substitute for
v
y
2
in equation (4):
(
)
2
0
2
0
2
2
0
2
0
4
3
2
1
y
x
y
x
v
v
v
v
+
⎟
⎠
⎞
⎜
⎝
⎛
=
+
Divide both sides by
v
0
x
2
and solve
for
v
0y
/
v
0x
to obtain:
1
16
9
2
1
1
2
0
2
0
2
0
2
0
⎟
⎟
⎠
⎞
⎜
⎜
⎝
⎛
+
=
+
x
y
x
y
v
v
v
v
⇒
7
0
0
=
x
y
v
v

Motion in One and Two Dimensions
229
Substitute for
x
y
v
v
0
0
in equation (1) and
evaluate
θ
0
:
(
)
°
=
=
⎟
⎟
⎠
⎞
⎜
⎜
⎝
⎛
=
−
−
3
.
69
7
tan
tan
1
0
0
1
0
x
y
v
v
θ
80
••
A cargo plane is flying horizontally at an altitude of 12 km with a
speed of 900 km/h when a large crate falls out of the rear-loading ramp. (Ignore
any effects due to air resistance.) (
a
) How long does it take the crate to hit the
ground? (
b
) How far horizontally is the crate from the point where it fell off when
it hits the ground? (
c
) How far is the crate from the aircraft when the crate hits the
ground, assuming that the plane continues to fly with the same velocity?

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- Staff
- Acceleration, Velocity, ΔT, Δv