Physics

An object is thrown upward with a speed of 21.5 m/s. How high above the projection point is it after 2.06 s?

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asked by James
  1. h = vₒ•t - g•t²/2

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    posted by Elena
  2. Thank you. This was really helpful. I was never given this equation.

    The only equation I got was:

    y= y_o + v_yo - 1/2(gt^2)

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    posted by James
  3. James. It is exactly the same equation.

    1/2 (a)= a/2

  4. Well, then I made a calculation error because the two equation gave me a different answer.

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    posted by James

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