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Projectile motion after a flying helical path

  • #1

Homework Statement


A bird flew along a helical path [tex]r(t)=<5cost,5sint,2t>[/tex], at time t=10 second, the bird died instantly. where did it hit the xy-plane ? g=32 ft/s/s.

Homework Equations


[tex]r(t)=(v_0 \cos\theta t) i+(v_0 \sin\theta t - 0.5gt^2 + h_0) j[/tex]

The Attempt at a Solution


taking derivative of r(t), I got the velocity function respect to t, [tex]v(t)=<-5sint,5cost,2>[/tex]
at t=10 s, we are located at (-4.2, -2.7, 20)
so that means the velocity in the x-direction at time =10 seconds is 2.72 ft/s
and the velocity in the y-direction at t=10 s = -4.2 ft/s
to hit the ground means that the distance in Z direction is 0, setting [tex]0=z(10)+v_z(10) t - 16t^2[/tex]
velocity in z direction is constant= 2
we can find the time the bird hit the ground, which = 1.18 second.
now, plug t and velocity in the x-direction into [tex]X_f=x(10)+v_x(10) t [/tex], we can find x-axis distance=-9.9 ft
[tex]Y_f=y(10)+v_y(10) t [/tex] similarly, we can find y - coordinate= -76.56 ft ( this value seems way off)

i am confused, does angle of projection matters in this case? if does, how do we find the angle of projection?
 
Last edited:

Answers and Replies

  • #2
BvU
Science Advisor
Homework Helper
2019 Award
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##
X_f=x(10)+v_x(10) t
= -4.2 + 2.7 * 1.18 ## can never be -9.9 ft !
For y you simply have a decimal point error, I'd guess: ##\ -2.7 - 1.18 * 4.2 = -7.7##

Always estimate the answer when using a calculator or a spreadsheet.

And I wouldn't round off before I had the answer.
 
  • #3
##
X_f=x(10)+v_x(10) t
= -4.2 + 2.7 * 1.18 ## can never be -9.9 ft !
For y you simply have a decimal point error, I'd guess: ##\ -2.7 - 1.18 * 4.2 = -7.7##

Always estimate the answer when using a calculator or a spreadsheet.

And I wouldn't round off before I had the answer.
so my approach was correct? thank you
 
  • #4
BvU
Science Advisor
Homework Helper
2019 Award
12,731
2,904
so my approach was correct? thank you
Approach was correct, execution a bit sloppy. And: You're welcome.
 

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