September30,2016
SectionB
MichelleLoven
ÜberRocketLab
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 37 degrees above the East horizontal. The rocket engine is designed to burn for 6.2 seconds
while producing a constant net acceleration of 4.1 m/s2 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 of 11 m/s. A horizontal wind blows the rocket, with the parachute, from the East to West at the constant speed of 20
m/s. Assume the wind affects the rocket only during the parachute stage.
444m/s
2
4.1
A)
Needed: Final coordinate position of A.
vf
The (sin) and (cos) relationships with
the hypothesis is used to find the
lengths of the X-dir and the Y-dir. This
gives the coordinates of the exact
moment when the rocket engine burns
out.
𝑣! = 𝑎 ∆𝑡 + 𝑣!
Y-dir: 78.802 sin 37
𝑣! = 4.1(6.2) + 0
X-dir: 78.802 cos 37
= 47.423 𝑚
= 62.9325 𝑚
B)
Needed: Final coordinate position of
B.
𝑣! = 25.42 m/𝑠 !
The Y-Max of B was found using Eq4:
!
!
𝑣!"#
= 𝑣!"#
+ 2 𝑎!" ∆𝑦
0 = (25.42 𝑠𝑖𝑛37 )! + 2 −9.8 ∆𝑦
Given:
θ = 37°
a = 4.1 m/s2
xi = 0 m/s
Δt = 6.2 s
yi
To find the distance traveled in A, Eq3
was used to find 𝑥! :
yf
−234.033 = −19.6 ∆𝑦
vi
∆𝑦 = 11.9405
∆𝑦 = 𝑦!" − 𝑦!"
𝑦!"# = 59.3605 𝑚
The parachute opened 44 m below max
height, so it opened at:
!
𝑥! = 𝑎(∆𝑡)! + 𝑣! ∆𝑡 + 𝑥!
!
xi
!
𝑥! = (4.1)(6.2)! + 0 6.2 + 0
!
𝑥 ! = 78.802 𝑚
Final velocity of A was calculated to
find initial velocity of B using Eq2,
because the moment the engine cuts off,
the projectile motion will begin.
Given:
xi = 0 m
yi = 47.423
xf
59.3605 𝑚 − 44 𝑚 = 15.3605 𝑚
𝑦! = 15.3605 𝑚
September30,2016
SectionB
Using Eq3, find the time it took for the
Y-dir to complete:
C)
Needed: final coordinate positions of C
!
MichelleLoven
Answer:
x-displacement of where the rocket lands
from the initial x-position
yi
𝑦!" = ! 𝑎!" ∆𝑡 ! + 𝑣!"# ∆𝑡 + 𝑦!"
15.36
!
= ! −9.8 ∆𝑡 ! + 25.42 𝑠𝑖𝑛37
∆𝑥 = 127.5 𝑚 East
∆𝑡
+ 47.423
yf
∆𝑡 = −1.4713𝑠 𝑜𝑟 4.55765𝑠
xi
xf
Plug this time into the X-dir equation,
and find 𝑥! , which represents the change
in distance of the hamster from the time
that the rocket engine shut off in the x
direction.
!
Given:
𝑥!" = 𝑥!" + 𝑥!" = 155.456 𝑚
𝑦! = 15.3605 𝑚
𝑦! = 0 𝑚
𝑣!" , 𝑣!" = 20𝑚/𝑠
𝑣!" , 𝑣!" = 11𝑚/𝑠
𝑥!" = ! 𝑎!" ∆𝑡 ! + 𝑣!"# ∆𝑡 + 𝑥!"
!
𝑥! = ! 0 4.55765 + 25.42 𝑐𝑜𝑠37
4.55765
+0
𝑥!" = 92.527 𝑚
The time change for the Y-dir vector was
calculated using Eq3:
!
𝑦! = ! 𝑎! ∆𝑡 ! + 𝑣!" ∆𝑡 + 𝑦!
!
Add this to the change in x position that
the rocket resulted in to get change in
distance of the hamster from the original
position.
0 = ! −9.8 ∆𝑡 ! + 11 ∆𝑡
92.527𝑚 + 62.933 𝑚 = 155.457 𝑚
∆𝑡 = −0.9738 𝑠 𝑜𝑟 1.3964 𝑠
+ 15.3605
15.36 = 11(∆𝑡)
Then, this time is plugged into the X-dir
equation Eq3 to get the xf of the windblown parachute:
!
𝑥! = ! 𝑎! ∆𝑡 ! + 𝑣!" ∆𝑡 + 𝑥!
!
𝑥! = ! 0 1.3964 + 20 1.3964 + 0
𝑥! = −27.93 𝑚
Add this to the change in x position that
the rocket resulted in to get change in
distance of the hamster from the original
position.
155.457 𝑚 − 27.93 𝑚 = 127.527 𝑚
This represents the total change in
distance in x of the entire voyage of the
hamster.