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Kinematics in Two Dimensions?

  • Thread starter pinky2468
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  • #1
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Kinematics in Two Dimensions??

Hi, I was wondering if anyone could give a little help on this problem. I can't figure out where to begin!
An Olympic long jumper leaves the ground at an angle of 23 degrees and travels through the air for a horizontal distance of 8.7m before landing. What is the take off speed of the jumper?
Thanks!!
 

Answers and Replies

  • #2
Tide
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Let x be horizontal and y be vertical and suppose the jumper takes off from the origin (0, 0) at t = 0. Then
[tex]x = v_0 \cos( \theta) t[/tex]
and
[tex]y = v_0 \sin(\theta) t - \frac{1}{2}g t^2[/tex]
so, first, find the time at which he/she lands (y = 0) and use that value of t in the first equation to find [itex]v_0[/itex]
 
  • #3
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Why is y=0 and for the y equation given is V=0?
 
  • #4
HallsofIvy
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pinky2468 said:
Why is y=0 and for the y equation given is V=0?
y=0 on landing because he is landing on the ground which where you are measuring from!

No, v0 is the same in both equations. (It is the sine and cosine that distinguish vertical and horizontal speeds.) That means that what Tide suggested isn't quite right- you can't "find the time at which he/she lands (y = 0) and use that value of t in the first equation to find v0".

What you can do is use the length of the jump given and solve the two simultaneous equations
[tex]v_0cos(23)t= 8.7 [/tex] and
[tex]v_0sin(23)t- \frac{g}{2}t^2= 0[/tex]
for both t and v0.
 
  • #5
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I'm sorry, I feel retarded but I still don't understand how you can have to unknown variables in one problem. If I don't know Vo or time how do I even solve the first equation?
 
  • #6
you can always remove one variable by substituting it as something else....

for instance, you can use t = x/(cos23 *v)
 
Last edited:
  • #7
Tide
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HallsofIvy said:
That means that what Tide suggested isn't quite right
Picky, picky! :smile:

Obviously, when I say "solve for" I meant solve for t in terms of [itex]v_0[itex].
 

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