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What is the final V in the vertical direction?

  1. Feb 13, 2019 at 11:55 PM #1
    1. The problem statement, all variables and given/known data

    2. Relevant equations
    Vy2 = Vyo2 + 2ayt
    y = 1/2(Vyo + Vy)t

    3. The attempt at a solution
    So I have "solved" this problem with both the equations above.

    Using the Vy2 equation I get the result of 32.83m/s.
    Using the other equation I get a result of zero, because Δy = (0 - 55) = -55. However, if I consider that 55m a positive value, then I get 32.83m/s... same as first equation.

    I knew Vyo = 0m/s because the initial velocity would only be in the x-direction since the ball "is thrown horizontally". To prove it I did the equation Δy = Vyot + 1/2ayt and got a result of 0m/s.

    So, where am I going wrong?

    I actually have a much bigger problem than just this one I'm posting. For some reason I cannot wrap my head around physics problems. I can do calculus, logic classes, philosophy, etc, but for some reason cannot make my mind look at physic's problems in the right way. Conceptually, all of these kinematic questions seem easy, but I am spending WAY more time on these problems than should be necessary. If you have an answer for this conundrum I welcome your responses...
  2. jcsd
  3. Feb 13, 2019 at 11:56 PM #2
  4. Feb 14, 2019 at 1:41 AM #3


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    I don't understand how you get ##V_y=0## using the second equation.

    Isnt it ##-55=\frac{1}{2}(0+V_y)3.35\Rightarrow V_y=\frac{-2\cdot55}{3.35}=-32.83##???
  5. Feb 16, 2019 at 3:26 PM #4


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    You have an error in the first equation you list
    Vy2 = Vyo2 + 2ayt​

    You should have Δy rather than t , so it should be:
    Vy2 = Vyo2 + 2ay(Δy) ,​
    which is probably what you actually used if you got 32.83m/s for Vy.

    Notice that when you solve an equation such as ##v_y^2 = 2(-9.8)(-55)## by taking the square root of both sides, the answer is:
    ##v_y = \pm \sqrt{1078} \approx \pm 32.83 ~. ##​

    You then need to choose the sign consistent with the situation
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