Über Problem: Hamster Huey and Algebra Alex
September 27, 2016
Nathan Ng
Section A
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 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:
Givens:
ymax
vwind
xi = 0 m
yCD = 70 m
ag = -9.8 m/s2
Θ = 34°
tAB = 8.0 s
vxwind = 16 m/s
aB = 6.6 m/s2
vywind = 10 m/s
Vxwind
Θ
xmax
Vywind
Strategy:
The strategy used to find the answer for the problem begins with listing out all of the givens. From there, the initial vertical and
horizontal velocities would need to be solved for. Next, the height at point B would need to be calculated, as well as the
distance traveled horizontally at point B. Then, from point B to point D, the rocket should be treated as a projectile. Then, the
rocket uses the vertical and horizontal velocities from point B as the velocities for the projectile motion throughout points A to
D. Using the equations of motion, the height from point Z to C can be found, which in turn can be added to the height found at
point B. This yields the max height of the entire projectile path of the rocket. Next, use the given distance at which the
parachute it released at and subtract it from the max height. Next, the distance traveled in the horizontal distance is needed to
be solved for. This will provide the x and y position of the rocket at point D. Next, using the release height, the time the
parachute takes to hit the ground can be solved for, by using the given vertical velocity of the wind. This time can then be
substituted into an equation of motion, along with the given horizontal velocity of the wind. When solved, this should yield the
distance traveled via wind. Going back to the equation of motion used for the projectile motion of the rocket, time for the
rocket from when the engine fails to when the parachute is released can be calculated. Using this time and the horizontal
velocity of the rocket at point B, the total distance from point B to point D is obtained. Using the distance traveled from point A
to point B, adding this value to the distance from point B to D provides the total horizontal distance traveled. Finally, subtract
the distance travelled via wind from the total distance travelled from point A to D, which will ultimately yield the answer.
Step 1: Finding the vertical position at point B
𝑦𝑍𝐶 = 44.49 𝑚
1
𝐵𝑦 = 𝑎𝐵 𝑡𝐴𝐵 2 + 𝑣𝑖𝑦 𝑡𝐴𝐵 + 𝑦𝑖
2
𝑦𝑚𝑎𝑥 = 𝑦𝑍𝐶 + 𝐵𝑦
1
𝐵𝑦 = (6.6𝑠𝑖𝑛34)(8.0 𝑠)2 + (0 𝑚/𝑠)(8.0 𝑠) + (0 𝑚)
2
𝑦𝑚𝑎𝑥 = (44.49 𝑚) + (118.1 𝑚)
𝑦𝑚𝑎𝑥 = 162.59 𝑚
𝐵𝑦 = 118.1 𝑚
Step 5: Find x and y position at point D
Step 2: Finding the x and y velocity components at point
B
1
𝑦𝐷 = 𝑎𝑔 𝑡𝐵𝐷 + 𝑣𝐵𝑦 𝑡𝐵𝐷 + 𝐵𝑦
2
𝑣𝐵𝑦 2 = 𝑣𝐴𝑦 2 + 2𝑎Δ𝑦
1
𝑚
𝑚
92.59 𝑚 = (−9.8 2 ) 𝑡𝐵𝐷 2 + (29.53 2 ) 𝑡𝐵𝐷 + 118.1 𝑚
2
𝑠
𝑠
𝑣 2 = (0 𝑚/𝑠)2 + 2(6.6𝑠𝑖𝑛34)(118.1 𝑚)
𝑣𝐵𝑦 = 29.53 𝑚/𝑠
𝑣𝐴𝑥
𝑣𝐵𝑦
=
𝑡𝑎𝑛34
𝑣𝐴𝑥 =
29.53 𝑚/𝑠
𝑡𝑎𝑛34
𝑣𝐴𝑥 = 43.78 𝑚/𝑠
Step 3: Finding the horizontal position at point B
𝑡𝐴𝐵 =
𝐵𝑦
𝐵𝑣𝑦
𝑡𝐴𝐵 =
118.1 𝑚
29.53 𝑚/𝑠
𝑥𝐵𝐷 = 𝑣𝐴𝑥 𝑡𝐵𝐷
𝑥𝐵𝐷 = 297.398 𝑚
Step 6: Find how the wind affects the rocket
𝒕𝒇𝑫 =
𝒚𝑫
𝒗𝒚𝒘𝒊𝒏𝒅
𝒕𝒇𝑫 =
𝟗𝟐. 𝟓𝟗 𝒎
𝟏𝟎 𝒎/𝒔
𝒕𝒇𝑫 = 𝟗. 𝟐𝟓𝟗 𝒔
𝑡𝐴𝐵 = 4.00
𝒙𝒘𝒊𝒏𝒅 = 𝒕𝒇𝑫 𝒗𝒙𝒘𝒊𝒏𝒅
𝒙𝒘𝒊𝒏𝒅 = (𝟗. 𝟐𝟓𝟗 𝒔)(𝟏𝟔 𝒎/𝒔)
𝒙𝒘𝒊𝒏𝒅 = 𝟏𝟒𝟖. 𝟏𝟒 𝒎
𝑥𝐴𝐵 = 𝑣𝐴𝑥 𝑡𝐴𝐵
𝑥𝐴𝐵 = 175.09 𝑚
Step 4: Find max height of rocket
2
𝑡𝐵𝐷 = 6.793 𝑠
Step 7: Find the final position
𝒙𝒎𝒂𝒙 = 𝒙𝑨𝑩 + 𝒙𝑩𝑫 − 𝒙𝒘𝒊𝒏𝒅
𝒙𝒎𝒂𝒙 = (𝟐𝟗𝟕. 𝟑𝟗𝟖 𝒎) + (𝟏𝟕𝟓. 𝟎𝟗 𝒎) − (𝟏𝟒𝟖. 𝟏𝟒 𝒎)
2
𝑣𝐶𝑦 = 𝑣𝐵𝑦 + 2𝑎Δ𝑦
(0 𝑚/𝑠)2 = (29.53 𝑚/𝑠)2 + 2(−9.8 𝑚/𝑠 2 )(𝑦𝑍𝐶 )
(0 𝑚/𝑠)2 = (29.53 𝑚/𝑠)2 + 2(−9.8 𝑚/𝑠 2 )(𝑦𝑍𝐶 )
𝒙𝒎𝒂𝒙 = 𝟑𝟐𝟒. 𝟑 𝒎