Homework 4
Due Tuesday 1/19/16
Barbarians hiding in a forest at attacking your village with a catapult. You can’t see them so you
are trying to locate them using nothing but a very accurate clock. The stone projectile is first seen
as they emerge from the forest at the top of the 8 meters tall trees (point A). Your measurements
show that it takes the stone 2.83 seconds to reach the edge of the forest 80 meters from your
village (point B). The stone arrives at your village 4.00 seconds after that (point C).
For both parts, assume that the ground is flat everywhere and that there is no air resistance.
B
A
8 m tall
trees
80 m to
edge of forest
barbarian
catapult
C
your
village
How far is the barbarian’s catapult from your village?
You set up a counter-catapult 30 meters from the edge of the forest. It launches stones at a
speed of 33 m/s. You also want your stones to come down on the barbarians’ catapult at the
steepest angle possible to avoid hitting the trees.
8 m tall
trees
barbarian
catapult
your counter-catapult
30 m to edge of forest
your
village
What angle should you launch your stones to hit the barbarian catapult?
page 1
In order to find information about the launch point, we need information about another point. That
point is going to be point A. However, we don’t know enough about point A in the beginning. We
do know something about points B and C, so here is the approach. Find information about point A
through points B and C then find information about the launch point.
Here is the description of the motion between points A and C.
launch point and set the time at launch to be zero.
Let me place the origin at the
B
A
y
8m
C
x
80 m
Here is the information connecting point A and point C. Note that the total flight time from A to C
is 6.83 s. We don't know what the time is a point A from the launch point but point C is 6.83
seconds after that. The velocity in the x direction is constant and it is the initial velocity times the
cosine of the launch angle. Another thing to note is that the velocity in the y direction at point C is
the negative of the initial velocity in the y direction and this is the initial velocity times the sine of
the launch angle.
x
a=0
tA
tA+6.83 s
xA
xC
vocosθ
vocosθ
y
a = –g
tA
tA+6.83 s
8m
0m
vA,y
–vosinθ
Here are the equations of motion.
xC − x A = vA,x (tC − tA )
vC ,y − vA,y = −9.8(tC − tA )
yC − yA = vA,y (tC − tA ) − 4.9(tC − tA )2
With values,
xC − x A = vo cos θ(6.83)
−vo sin θ − vA,y = −9.8(6.83)
−8 = vA,y (6.83) − 4.9(6.83)2
The x direction doesn’t help at all. The y direction, however, gives us this.
page 2
−8 = vA,y (6.83) − 4.9(6.83)2
⇒ vA,y = 32.296
−vo sin θ − vA,y = −9.8(6.83) ⇒ vo sin θ = 34.638
This is not the answer yet but we are close. Let’s look at the information from point B to point C.
This takes 4.00 seconds and the stone travels 80 meters horizontally. Again, the final velocity in
the y direction is negative of the initial velocity in the y direction.
x
a=0
tB
tB+4.00 s
xB
xB+80 m
vocosθ
vocosθ
tB
yB
vB,y
y
a = –g
tB+4.00 s
0m
–vosinθ
Here are the equations of motion.
xC − x B = vB,x (tC − tB )
vC ,y − vB,y = −9.8(tC − tB )
yC − yB = vB,y (tC − tB ) − 4.9(tC − tB )2
With values,
80 = vo cos θ(4.00)
−vo sin θ − vB,y = −9.8(4.00)
yC = vB,y (4.00) − 4.9(4.00)2
In this case, the y direction doesn’t help at all. The x direction, however, gives us this.
80 = vo cos θ(4.00) ⇒ vo cos θ = 20
Taking the two pieces of information arrived at through each segment of time together, I will divide
one by the other.
vo sin θ
34.638
=
vo cos θ
20
⇒
tan θ = 1.7319
⇒ θ = 60.0°
The initial speed is
vo cos 60° = 20
⇒ vo = 40
Here is the launch profile of the barbarian catapult: 40 m/s at 60°. This means the range of this
shot is
R=
vo2
402
sin(2θ) =
sin(2 ⋅ 60°) = 141.39
g
9.8
The barbarian catapult is 141 meters from the village.
page 3
Your counter-catapult being 30 meters from the edge of the forest means you are 91 meters from
the barbarians. Here is my coordinate system.
y
x
91 m
Here are the information about my stone’s trajectory.
x
a=0
0
0
vocosθ
y
a = –g
t
91.39 m
vocosθ
0
0
vosinθ
t
0
–vosinθ
Here are the equations of motion.
91.39 = 33cos θ ⋅ t
−2 ⋅ 33sin θ = −9.8 ⋅ t
0 = 33sin θ ⋅ t − 4.9 ⋅ t 2
Here are the solutions.
0 = 33sin θ ⋅ t − 4.9 ⋅ t 2
t = 0 and t =
⇒ t(33sin θ − 4.9 ⋅ t) = 0
33
sin θ
4.9
Using the second solution, after t = 0, for the flight time,
91.39 = 33cos θ ⋅ t
⇒
91.39 = 33cos θ ⋅
33
sin θ
4.9
⇒
2sin θ cos θ = sin(2θ) = 0.82244
The solution to 2θ are the following angles as the sine function is positive in the first and second
quadrants.
page 4
sin(2θ) = 0.82244
⇒
2θ = 55.33°, 124.67°
The angles are
θ = 27.7°, 62.3°
The steeper launch and landing angle is 62.3°.
page 5