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Hwk Question

  1. Nov 6, 2007 #1
    A Boy is standing at the edge of a cliff. He holds his hand over the edge of the cliff and throws a pebble directly up into the air with an initial speed of 11.6 m/s.
    a) How long does it take for the pebble to reach maximum height?
    b) What is the pebbles velocity when it hits the base of the cliff exactly 7.16 seconds after it was thrown? What is the pebble's speed at the same point in time?

    average speed= distance/time; average velocity= (VF+VI)/2; average velocity= displacement in position/change in time; average acceleration= change in velocity/change in time; change in position= 1/2at^2
  2. jcsd
  3. Nov 6, 2007 #2


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