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Maths

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A ball bounces two-thirds of the distance it falls. If it is dropped from a height of 10 meters, how far does it move before hitting the floor for the fourth time?

  • Maths -

    Distance traveled :
    first bounce -- 10 m
    2nd bounce -- (2/3)(10) or 20/3 both up and down
    3rd bounce -- (2/3)(20/3) or 40/9 both up and down
    4th bounce -- (2/3)(40/9) or 80/27 both up and down

    total distance
    = 10 + 2(20/3) + 2(40/9) + 2(80/27)
    = 886/27 m
    or appr 32.8 m

  • Maths -

    hi Reiny,
    U solved absolutely wrong, plz Ali follow this method
    Solution:
    For Downward Distance:-
    a=10 (first term)
    r=2/3 (common ratio)
    n=4 (no. of terms)
    s=? (sum of 4th terms)
    Sn= a(1-r^n)/1-r (because r is less than1)
    S= 10(1-(2/3)^4/1-(2/3)= 650/27

    For Upward Distance
    a=10*2/3=>20/3
    r=2/3
    n=3
    S=?
    S=a(1-r^n)/1-r
    S=20/3(1-(2/3)^4/1-(2/3)
    S=380/27
    Now, add downward distance & upward distance for calculating total distance

    650/27 + 380/27 = 1030/27

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