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Components of velocity

  1. Oct 12, 2007 #1

    klm

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    A ball is thrown toward a cliff of height h with a speed of 30m/s and an angle of 60 degrees above horizontal. It lands on the edge of the cliff 4.0s later.

    How high is the cliff?
    do you just use the equation h=.5gt^2 ? and put it in -9.80=g and 4=t ?
     
  2. jcsd
  3. Oct 12, 2007 #2

    Doc Al

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    Staff: Mentor

    You also need to consider the initial speed in the y-direction; here's the complete equation:
    [tex]y = y_0 + v_{0y} t - (1/2) g t^2[/tex]
     
    Last edited: Oct 12, 2007
  4. Oct 12, 2007 #3

    klm

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    would this be correct: y0=26 , v0=30, g=-9.8 , t=4 ... so y=26+(30x4)- .5(-9.8)(4)^2 = 224.4 m ?

    i got vy by doing v0sintheta= 26
     
  5. Oct 12, 2007 #4

    Doc Al

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    Staff: Mentor

    No. In the formula I gave:
    y0 is your initial position, which I presume is on the ground at height = 0
    v0 (which I'll change to v0y) is the vertical component of the initial velocity, what you call v0sin(theta)
    g = 9.8 m/s^2

    To make it less confusing, I'll relabel v0 to be v0y in my equation.
     
  6. Oct 12, 2007 #5

    klm

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    ohh sorry, so y=0 + 26(4) - .5(9.8)(4)^2 = 25.6 m
     
  7. Oct 12, 2007 #6

    Doc Al

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    Good!
     
  8. Oct 12, 2007 #7

    klm

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    thanks! can you help me with the next part too..! What was the maximum height of the ball?
    i think the equation is just the same as the one you wrote, but just cut t=4 in half to get the peak height so t=2 ..so y= 26(2)-.5(9.8)(2)^2 = 32.4 m ?
     
  9. Oct 12, 2007 #8

    Doc Al

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    No, you can't assume that the peak is at half the time. After all, it lands up on a cliff, so it spent more time rising than falling. (If it fell back down to the original height, then you'd be correct.)

    Instead, use a velocity equation for the y-direction to figure out the time when it reaches maximum height. Hint: At the maximum height, what's the vertical speed?
     
  10. Oct 12, 2007 #9

    klm

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    um i think the vertical speed should be 0
     
  11. Oct 12, 2007 #10

    klm

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    so would it be okay to use the equation vfy= viy +ayT so 0= 26+9.8t so t= 2.65 and then cut then stick that time in that first equation you gave me?
     
  12. Oct 12, 2007 #11

    klm

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    so y= 26(2.65) -.5(9.8)(2.65)^2 =34 .4m ?
     
  13. Oct 12, 2007 #12

    klm

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    are the equations i used alright?
     
  14. Oct 12, 2007 #13

    Doc Al

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    Very good!
     
  15. Oct 12, 2007 #14

    klm

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    thank you Doc Al! do you mind one more question, it will be the last one i promise! =)
     
  16. Oct 12, 2007 #15

    klm

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    What is the ball's impact speed?

    i thought what you should do is find the final velocity in the x component and y component. and i thought that vfx= 15 since there is no acc in the x direction and vfy= -26 b/c of neg acc. and then i thought you should take the magnitude, but this does not work out to be right. do you know what i am doing wrong

    wait i think i did this wrong, look at my next post please
     
    Last edited: Oct 12, 2007
  17. Oct 12, 2007 #16

    klm

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    oh no actually should vfy= 65.2 because i tried the equation vfy=viy+ay x T so 26+(9.8x4)= 65.2
    so do i do square root (15^2 + 65.2^2) = 66.9 ?
     
  18. Oct 12, 2007 #17

    Doc Al

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    Careful here. ay = -9.8 m/s^2.
     
  19. Oct 12, 2007 #18

    klm

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    ohhh so should it be vfy= -13.2 and then do the sqaure root (13.2^2 +15^2 ) =19.98
     
  20. Oct 12, 2007 #19

    Doc Al

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    Good!
     
  21. Oct 12, 2007 #20

    klm

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    thank you so much doc al!!
     
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