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Qck. quest. needs qck. answ.

  1. Sep 14, 2007 #1
    In its final trip upstream to its spawwning territory, a salmon jumps to the top of a waterfall 1.9m high. What is the minimum vertical velocity needed by the salmon to reach the top of the waterfall?

    Is there enough given in this problem to solve it???
    All I can see is; ay=-9.8m/s^2 and dy= 1.9m.
    What do you think???
  2. jcsd
  3. Sep 14, 2007 #2


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    What will the final vertical velocity of the salmon be if we are interested in the minimum initial vertical velocity?
  4. Sep 14, 2007 #3
    While the only thing I can come up with is that the minimun velocity must be just enough to overcome gravity. But I still don't know how to prove this
  5. Sep 14, 2007 #4


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    Yes, that's the minimum velocity that the question is asking for. But, for that minimum velocity there is something "special" about the final vertical velocity of the salmon when it reaches the top of the waterfall. Think of this as a projectile. The minimum initial vertical velocity will occur when the peak of the projectile is at the exact point at the top of the waterfall. What do you know about the vertical velocity of an object at the peak of a parabolic trajectory?
  6. Sep 14, 2007 #5
    AHHH! The final velocity will be zero because it has reached the point where it and gravity are in equilibrium.
  7. Sep 14, 2007 #6


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    The final vertical velocity is indeed zero. With this you should be able to solve the problem using one of the kinematic equations.
  8. Sep 14, 2007 #7
    Thank you very much, I appreciate you help. Quick and concise...PERFECT!
  9. Nov 11, 2007 #8
  10. Nov 11, 2007 #9
    i got the same question...and came up with an answer that doesn't make sense.

    i got

    Vf^2 = Vi^2 + 2 (a) (d)

    0 = Vi^2 + 2(-9.8m/s^2) (1.9m)

    -Vi^2 = -18.6m/s^2 (1.9m)

    Sqroot (-Vi^2) = Sqroot (-35.34 m^2/S^2)

    Vi = 5.9 m/s
  11. Nov 11, 2007 #10
    how can Vi be 5.9 m/s when gravity is -9.8 m/s?
  12. Nov 11, 2007 #11


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    What are you bumping for, jon? If you have problems with this question then post your work and we will help you.
  13. Nov 11, 2007 #12
    oh...I just added couple posts...

    and btw....I like how you get people to find the answer by cluing them in, instead of just telling them the answer. Cool!!
  14. Nov 11, 2007 #13
    g is NOT a velocity!!! Look at your units for g, they are incorrect.

  15. Nov 11, 2007 #14

    what is incorrect?

    that math itself? so .... 0 = Vi^2 + 2(-9.8 m/s^2) (1.9m) is correct but I'm doing the math wrong?
  16. Nov 11, 2007 #15
    He's saying that g is acceleration, not velocity, and so it should have units of m/s^2.

    I'd say your math is correct. velocity can be in any direction, it does not require that it be in the same direction as acceleration. If they have opposite signs, it just means that the object is slowing down.

    Also, You should have dropped the negatives before square-rooting, or left them outside of the root sign. You can't square root a negative number.
  17. Apr 30, 2008 #16
    I am doing the same question, and just want to double check if I got the math correct and the correct answer:
    0 = Vi^2 + 2(-9.8m/s^2) (1.9m)
    0= Vi^2+(-37.24)
    Vi=sqrt 37.24
    vi=6.10 m/s (is this correct?)
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