Uber Problem: Hamster Huey and Algebra Alex
Philip Economou
Section C
September 30, 2015
Description
One breezy afternoon Algebra Alex decides to launch Hamster Huey into the air using a model rocket. The rocket is launched over level
ground, from rest, at a specified angle above the East horizontal. The rocket engine is designed to burn for specified time while producing
a constant net acceleration for the rocket. Assume the rocket travels in a straight-line path while the engine burns. After the engine stops
the rocket continues in projectile motion. A parachute opens after the rocket falls a specified vertical distance from its maximum height.
When the parachute opens the rocket instantly changes speed and descends at a constant vertical speed. A horizontal wind blows the
rocket, with parachute, from the East to West at the constant speed of the wind. Assume the wind affects the rocket only during the
parachute stage.
Diagram
Work
1 - Leg AB:
B
ay
𝐴𝐵 =
viB
∆tAB
aAB
viA
xi 45°
A
viCx C
B ΘA
ymax
yiB
viCy
xiB
xf
𝐴𝐵 =
1
(6.7)(7.9)2 + 0 + 0
2
𝐴𝐵 = 209.074 𝑚
yiC
𝑥𝑖𝑏 = 𝐴𝐵𝑐𝑜𝑠(45)
D
A ΘA
xiB
1
𝑎 𝑡 2 + 𝑣𝑖𝐴 𝑡 + 𝑥𝑖
2 𝐴𝐵
xiC
𝑥𝑖𝑏 = 209.074cos(45)
𝑥𝑖𝐵 = 147.837 m
Givens
sin(45) = cos(45),
Since
viA = 0 m/s
yi = 0 m
xi = 0 m
ΘA = 45 degrees
∆tAB = 7.9 sec
aAB = 6.7 m/sec2
xiB = ?
yiB = ?
ymax - yic = 87m
viB = ?
ay = -9.8 m/sec2
ymax = ?
viCx = -14 m/s
𝑥𝑖𝐵 = 𝑦𝑖𝐵
viCy = -6 m/s
yiC = ?
𝑣𝑖𝐵 = 𝑎𝐴𝐵 ∆𝑡𝐴𝐵 + 𝑣𝑖𝐴
𝑦𝑖𝐵 = 147.837 𝑚
Calculating ViB:
𝑣𝑖𝐵 = (6.7)(7.9) + 0
xic = ?
xf = ?
𝑣𝑖𝐵 = 52.93 𝑚/𝑠
2 - Leg BC:
𝑣𝑖𝐵𝑥 = 𝑣𝑖𝐵 cos(45)
Strategy:
1.
2.
3.
Treat AB as a linear motion problem (do not
componentize). Solve for characteristics of point B (x
and y position and velocity) Angle at point B is equivalent
to ΘA, since the rocket traveled in a straight line.
Solve for ymax, using equation for y during AB in terms
of time. Subtract 87 to find yiC. Find time at yic,
substitute into x equation to find xic.
Directional shift occurs instantly, meaning that
finding the theoretical initial velocity based on vector
equations that apply to AB is unnecessary. Create x and
y equations for the final leg from the given values of vicy
and vicx. Set y equal to zero, solve for t. Substitute t into
x equation. Since x and y positions have been carried
through all parts of the problem, this directly yields xf.
viB
B
𝑣𝑖𝐵𝑥 = 52.93 cos(45)
viBy
45°
viBx
𝑣𝑖𝐵𝑥 = 37.4272 𝑚/𝑠
𝑣𝑖𝐵𝑦 = 𝑣𝑖𝐵𝑥 (see above)
𝑣𝑖𝐵𝑦 = 37.4272 𝑚/𝑠
Using this information to express the rockets’ x and y positions
during leg AB in terms of time,
𝑥[𝑡] =
1
𝑎 𝑡 2 + 𝑣𝑖𝐵𝑥 𝑡 + 𝑥𝑖𝐵
2 𝑥
𝑥[𝑡] = 0 + 37.4272𝑡 + 147.837
𝑥[𝑡] = 37.4272𝑡 + 147.837
𝑦[𝑡] =
1
𝑎 𝑡 2 + 𝑣𝑖𝐵𝑦 𝑡 + 𝑦𝑖𝐵
2 𝑦
𝑦[𝑡] = (0.5)(−9.8)𝑡 2 + 37.4272𝑡 + 147.837
𝑦[𝑡] = −4.9𝑡 2 + 37.4272𝑡 + 147.837
In the general form of the quadratic equation, 𝑦 = 𝐴𝑥 2 + 𝐵𝑥 +
𝐵
𝐶, the expression − , gives the x-coordinate of the vertex.
2𝐴
When applied to y[t], we obtain the time at the maximum height,
which we can substitute into y[t] to yield the height itself.
𝐵
𝑡𝑦𝑚𝑎𝑥 = −2𝐴
𝑡𝑦𝑚𝑎𝑥 = −
37.4272
2(−4.9)
𝑡𝑦𝑚𝑎𝑥 = 3.8191
𝑦[3.8191] = −4.9(3.8191)2 + 37.4272(3.8191) + 147.837
𝑦𝑚𝑎𝑥 = 219.306
From the given information,
𝑦𝑖𝐶 = 𝑦𝑚𝑎𝑥 − 87
𝑦𝑖𝐶 = 219.306 − 87
𝑦𝑖𝐶 = 132.306
By setting y[t] equal to this quantity, we can obtain the time at
which the rocket is at point C. Substituting the result into x[t]
yields the x coordinate at this time, xiC.
132.306 = −4.9𝑡 2 + 37.4272𝑡 + 147.837
−4.9𝑡 2 + 37.427𝑡 + 15.531 = 0 (solver)
𝑡 = −.394584 , 8.03275
Substituting into x[t],
𝑥[8.03275] = 37.4272(8.03275) + 147.837
𝑥𝑖𝑐 = 448.48
3 – Leg CD
Creating x and y position equations between C and D based on
given x and y constant velocities,
𝑥[𝑡] =
1
𝑎 𝑡 2 + 𝑣𝑖𝑐𝑥 𝑡 + 𝑥𝑖𝐶
2 𝑥
𝑥[𝑡] = 0 − 14𝑡 + 448.48
𝑥[𝑡] = −14𝑡𝑡 + 448.48
𝑦[𝑡] =
1
𝑎 𝑡 2 + 𝑣𝑖𝑐𝑦 𝑡 + 𝑦𝑖𝐶
2 𝑦
𝑦[𝑡] = 0 − 6𝑡 + 132.306
𝑦[𝑡] = −6𝑡 + 132.306
Setting y[t] to zero to find the time at which the rocket hits the
ground,
−6𝑡 + 132.306 = 0
−6𝑡 = −132.306
𝑡 = 22.051
Substituting into x[t] to find the x position at this time, x f,
𝑥[22.051] = −14(22.051) + 448.48
𝑥𝑓 = 139.8 𝑚