Calculus

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.

  1. 👍 0
  2. 👎 0
  3. 👁 232
  1. 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

    1. 👍 0
    2. 👎 0
  2. 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?

    1. 👍 0
    2. 👎 0

Respond to this Question

First Name

Your Response

Similar Questions

  1. science

    a ball is thrown straight upward and returns to the thrower's hand after 3.00s in the air. A second ball is thrown at an angle of 30.0 deg with the horizontal. At what speed must the second ball be thrown so that it reached the

  2. Calculus

    If a ball is thrown vertically upward from the roof of 32 foot building with a velocity of 112 ft/sec, its height after t seconds is s(t)=32+112t–16t2 . What is the maximum height the ball reaches?What is the velocity of the

  3. Physics

    A ball is thrown horizontally from the top of a building. The ball is thrown with a horizontal speed of 8.2 ms^-1. The side of the building is vertical. At point P on the path of the ball, the ball is distance x from the building

  4. Math

    A rubber ball is dropped from the top of a building, which is 40 ft. above the ground. Each time the ball hits the sidewalk, it rebounds 77% of its previous height. How high will the ball rebound after its sixth bounce? Round your

  1. calculus

    If a ball is thrown vertically upward from the roof of 64 foot building with a velocity of 64 ft/sec, its height after t seconds is s(t)=64+64t–16t2 . What is the maximum height the ball reaches? What is the velocity of the ball

  2. Physics

    A ball is thrown horizontally at 18m/s from the roof of a building lands 28metre from the base of the building. Calculate the height of the building.

  3. College Calculus

    If a ball is thrown vertically upward from the roof of 48 foot building with a velocity of 96 ft/sec, its height after t seconds is s(t)=48+96t−16t^2. What is the maximum height the ball reaches? What is the velocity of the ball

  4. Quadratic equations

    A ball is thrown vertically upwards From the top of a building of height 29.4 m and with an initial velocity 24.5 m/sec. If the height H of the ball from the ground level is given by H = 29.4 + 24.5t - 4.9t², then find the time

  1. Dynamics

    A 70-g ball B dropped from a height h_0=1.5 m reaches a height h_2=0.25 m after bouncing twice from identical 210-g plates. Plate A rests directly on hard ground, while plate C rests on a foam-rubber mat. Determine (a) the

  2. College Algebra

    Flight of a Ball If a ball is thrown upward at 96 feet per second from the top of a building that is 100 feet high, the height of the ball can be modeled by S(t)=100 + 96t -16t^2 feet, where t is the number of seconds after the

  3. phy

    A blue ball is thrown upward with an initial speed of 20.6 m/s, from a height of 0.8 meters above the ground. 2.5 seconds after the blue ball is thrown, a red ball is thrown down with an initial speed of 8.2 m/s from a height of

  4. math

    a ball is thrown upward at 64 feet per second from the top of an 80 feet high building. The height of the ball can be modeled by S(t) = -16t^2 + 64t + 80(feet), where t is the number of seconds after the ball is thrown. describe

You can view more similar questions or ask a new question.