physics

a 2.50 N rock is thrown down from a cliff at 10.0 m/s.
a. what is the velocity after falling 25.0 m?
b. velocity as it hits the water 125 m below?
c. kinetic energy as it hits the H20?

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asked by carol
  1. I am not going to do all of these for you. This and the previous one are basically the same.

    The mass does not matter until we get to part c

    define velocity and distance positive down
    Vi = 10
    g = 9.81
    distance at 0 = 0
    v = Vi + g t
    d = Vi t + (1/2) g t^2
    at 25 meters
    25 = 10 t + 4.9 t^2
    4.9 t^2 + 10 t -25 = 0
    t = [ -10 +/- sqrt(100+ 490)]/9.81
    t = [-10 +/- 24.3 ]/9.81
    t = 1.46 seconds
    so
    v = 10 + g t = 24.3 m/s at 25 m

    at water
    d = 125
    125 = 10 t + 4.9 t^2
    4.9 t^2 + 10 t - 125 = 0
    t = [ -10 +/- sqrt (100+2450)]/9.81
    t = 4.13 seconds
    so
    v = 10 + 9.81(4.13) = 50.5 m/s at 125 m

    Ke = (1/2) m v^2
    = (1/2)(2.5/9.81)(50.5)^2
    = 325 Joules

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    posted by Damon

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