Calculus

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Hard rubber ball is thrown down from a building with a height of 5 metres. After each bounce, the ball rises to 4/5 of its previous height. Total vertical distance d ball has travelled at the moment it hits the ground for the 8th time, to the nearest 10th of a metre, is..____________

I don't get how to solve this.

  • Calculus -

    make a sketch and you will see that the total distances
    = 5 + 2(5)(4/5) + 2(5)(4/5)^2 + .... for 8 terms
    = 5 + 10(4/5) + 10(4/5)^2 + ... for 8 terms

    notice that the first term is not part of the pattern since it does not complete the "up-and-down" path
    lets make it fit the patters:
    total distance
    = (10 + 10(4/5) + 10(4/5)^2 + ... 10(4/5)^7) - 5
    = 10 ((4/5)^8 - 1)/(4/5-1) - 5
    = 36.6

    check:
    1st bounce = 5
    2nd bounce = 8 , 4 up and 4 down
    3rd bounce = 6.4
    3rd bounce = 5.12
    4th bounce = 4.096
    5th bounce = 3.2768
    6th bounce = 2.62144
    7th bounce = 2.097152
    8th bounce = 1.6777216
    sum of those 8 terms = 36.611392

  • Calculus -

    A ball with mass 0.15 kg is thrown upward with initial velocity 20 m per sec from a roof of a building 30 m high find the max. height a ball reach?

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