Question:

Free fall acceleration?

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When stratled, an armadillo will leap upwards. Suppose it was 0.544m in the first 0.200s(a) What is its initial speed as it leaves the ground/(b) what is its speedat the height of 0.544m?(c) How much higher does it go?

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  1. Given:

    at t = 0.200 s, d = 0.544m

    Find:

    a)  Vi, initial speed

    b)  Vf, speed at height 0.544m

    c)  h = Dmx - 0.544, how much higher it go

    Solution:

    a)  d = Vit + (1/2)gt^2

         0.544 = Vi(0.200) + (1/2)(-9.8)(0.200)^2

         Vi = [0.544 + (1/2)(9.8)(0.200)^2]/0.200

         Vi = 3.7 m/s    ANSWER

    b)  Vf = Vi + gt

          Vf = 3.7 + (-9.8)(0.200)

          Vf = 1.74 m/s      ANSWER

    c)   2gDmax = Vf^2 - Vi^2

               Dmax = (0^2 - 3.7^2)/(2)(-9.8)

               Dmax = 0.70 m

          h = Dmax - d

             = 0.698 m - 0.544 m

             = 0.154 m    ANSWER

    Hope this helps.

    teddy boy

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