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A ball is shot from the ground into the air

  1. Sep 18, 2016 #1
    1. The problem statement, all variables and given/known data
    A ball is shot from the ground into the air. At a height of 8.8 m, the velocity is observed to be

    v = (7.4)i + (5.7)j in meters per second (i horizontal, j upward). To what maximum height will the ball rise?

    2. Relevant equations
    V^2 = Vi^2 + 2a(Δy)

    3. The attempt at a solution

    I have listed my knowns for x and y
    Y:
    Yi = 0m
    Y = 8.8 m
    V = 5.7 m/s
    Vi = 11.8 m/s ~solved from the equation

    X:
    V = 7.4 m/s
    Xi = 0m

    I am having a difficult time finding the max height because they do not give the time and I am not sure if I could solve for time with only the y components.

     
  2. jcsd
  3. Sep 18, 2016 #2

    billy_joule

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    Science Advisor

    You don't need the time.
    You can use the same equation again.
    What is the y velocity when max height is reached?
     
  4. Sep 18, 2016 #3
    The velocity would be zero but I used that number and I get 7.1 m for displacement and that's too low.
     
  5. Sep 18, 2016 #4
    I see what I was doing wrong. I was forgetting to put in the 8.8 m and the 5.7 velocity which they already gave to me. Instead, I was using the initial velocity that I found which was not necessary.
     
  6. Sep 18, 2016 #5

    billy_joule

    User Avatar
    Science Advisor

    As long as you recognise the starting points are different you can use either velocity, the answer is the same.
     
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