college physics

posted by tina

While standing on a bridge 15.0 m above the ground, you drop a stone from rest. When the stone has fallen 3.10 m, you throw a second stone straight down. What initial velocity must you give the second stone if they are both to reach the ground at the same instant? Take the downward direction to be the negative direction

  1. tina

    (-15.0)-(1/2)(-9.80)(.96)/.96 answer equals .10.7

  2. Damon

    Tina - I did this problem for Brittany, scroll down.

  3. Damon

    first get time to fall to ground
    a = -9.8
    v = -9.8 t
    -15 = -4.9 t^2
    t = sqrt (15/4.9) = 1.75 seconds
    -------------------------------
    Now how long did it take to reach 3.1 m down
    3.1 = -4.9t^2
    t = .795 seconds
    ---------------------------------
    so the second rock spends 1.75 -.795 or .955 seconds in the air
    -15 = Vo (.955) -4.9 (.955)^2
    .955 Vo = 3.64-15
    Vo = - 11.9
    so 11.9 m/s downward (negative)

  4. Camrik

    wrong answer -4.9(.955)^2 does not equal 3.64

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