2D Kinematics On Planet Zorg

  • #1
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Hello everybody. Although this is just a kinematics problem, im having a hard time figuring out what to do. The question is:

An archer on the planet zorg spots an octomorph that is resting in a tree above water. The octomorph's position in the tree is 100m away horizontally and 100m vertically away from the bow of the archer. The archer shoots a quantum arrow at the creature with an initial speed of 100m/s, however, at the instant the arrow is fired, the octomorph drops straight out of the tree into the water below. If the acceleration of gravity on planet zorg is 98.1m/s^2 , in what direction should the archer aim his arrow if it is to hit the octomorph in stride on the way down?



Homework Equations


y= -1/2g(t^2)+Voy(t) + yo
x= Vxt

The Attempt at a Solution


I have figured out the position of the octomorph as a function of time : y=(-49.05m/s^2)t^2 + 100m

I have no idea how to make the arrow and octomorph reach the same position in the same time. We cannot simply set their equations equal to eachother as the arrow has an x component to take into consideration. Any help or ideas will be greatly appreciated!
 

Answers and Replies

  • #2
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You are going to have two equations in two unknowns (the time and the angle). The horizontal location of the arrow has to match the horizontal location of the octomorph, and the vertical location of the arrow has to match the vertical location of the octomorph. Let ##\theta## be the angle (measured from the horizontal) that the arrow is launched. In terms of ##\theta##, what are the x and y components of the arrow velocity? In terms of ##\theta## and t, what are the horizontal and vertical locations of the arrow at time t (relative to the archer)?

Chet
 
  • #3
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The horizontal component of velocity is (Vo)cosθ which is (100m/s)cosθ

And the vertical component of velocity will be (100m/s)sinθ

The position equation in the horizontal direction will be

X = (100m/s)cosθ(t)

And for the y direction it will be

Y = (-49.05m/s^2)t^2 + (100m/s)sinθ(t)
 
  • #4
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Im thinking now we might have to combine and somehow manipulate these equations in order to solve for the angle and time. Not sure exactly how since there are two unknowns. Thats what im thinking atleast.. but i will wait for your response
 
  • #5
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Also i believe that the value for x will be 100m regardless of what occurs in the y direction. Meaning that the full equation for the horizontal direction will be

100m = (100m/s)cosθ(t)
cosθ = 1s(second)/t
θ = arcos( s/t )

is that right?
 
  • #6
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How are the x and y coordinates of the arrow related to the x and y coordinates of the octorph when the arrow hits the octomprph?

Chet
 
  • #7
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How are the x and y coordinates of the arrow related to the x and y coordinates of the octorph when the arrow hits the octomprph?

Chet
The will both have the same x and y coordinates.

Both x coordinates will be at X = 100m

I think also means that the y equations will be equal to each other.

so
(-49.05m/s^2)t^2 +(100m/s)sinθ(t) = (-4.905m/s^2)t^2 +100m
 
  • #8
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So, you can cancel the acceleration terms from both sides of the y equation. What does that leave you?

Chet
 
  • #9
epenguin
Homework Helper
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I don't think any equations or calculations are needed to answer this question.
 
Last edited:
  • #10
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I don't think any equations or calculations are needed to answer thus question.
I don't think so either, but that is only because we both have lots of experience. I don't think I would have known this as a new initiate like the OP. (Probably you would have).

Chet
 
  • #11
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So, you can cancel the acceleration terms from both sides of the y equation. What does that leave you?

Chet
Yes they can be canceled.

This will leave the equation as

(100m/s)sinθ(t) = 100m

Im sorry.. but still cant see where this is going. Sorry for the late reply. i had to attend some classes
 
  • #12
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So you have:

100 sinΘ t = 100

and

100 cosΘ t =100

What do you get if you divide the first equation by the second?

Chet
 
  • #13
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So you have:

100 sinΘ t = 100

and

100 cosΘ t =100

What do you get if you divide the first equation by the second?

Chet

You divide them so
tanθ=1
θ=arctan(1)=45 degrees

Is that right
 
  • #15
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