Projectile motion - Kinematics

  • Thread starter rejz55
  • Start date
  • #1
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Homework Statement



At 75% its maximum height, the speed of a projectile is four-fifths its initial speed. What is the angle of the initial velocity vector with respect to the vertical?

I really have no idea how to do this. Can you give me the idea on how to start or how to do it. (just please don't tell me the answer)

Thanks a lot!
 

Answers and Replies

  • #2
Shooting Star
Homework Helper
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At the max height, there is no component of velo in y dircn. vx remains constant. vy is given by the eqn vy^2 = vy_0^2 - 2gh.

Do you understand something now?
 
  • #3
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if i will use that equation there will be 2 unknown variables. the initial velocity of y and the height..

how can i use the given 75% its max height and the 4/5 speed of the projectile
 
  • #4
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As shooting star wrote,

[tex]{v_{y0}}^2 = 2gy_\textrm{max}[/tex]

Try working with that equation and

[tex]{v_{y2}}^2 = {v_{y0}}^2 - 2gy_{2}.[/tex]

Hint: Remember your givens.
 
Last edited:
  • #5
Shooting Star
Homework Helper
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how can i use the given 75% its max height and the 4/5 speed of the projectile

You only have to find the ratio of vx/vy_0, not the absolute values.
 
  • #6
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Thanks!! Got it!
 

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