physics

A 8kg ball is thrown up with an initial speed of 20m/s. Suppose there is a constant air resistence of 30N opposing the motion during its entire up and down path. How long will it take to get to the highest point in its path? how high is the highest point? How much work was done by the air resistence during the entire path up and down? What is the kinectic energy when it gets back to its original point?

asked by dipset
  1. I will be glad to critique your work.

    posted by drwls
  2. I don't how to start it off

    posted by dipset
  3. When the ball reaches its highest point H, potential energy will be
    M g H = (1/2)MVo^2 - f*H
    where f is the friction force, 30N.

    H = (1/2)MVo^2/(Mg + f)

    Solve for H.

    The time to get to that high point will be
    T = H/Vo, since it will be decelerating at a constant rate.

    The KE when it hits the ground will be M g H - f*H. This will equal
    (1/2)MVo^2 - 2fH

    posted by drwls
  4. Thank you. one more question what would be the acceleration on the way up? on the way down?

    posted by dipset
  5. acceleration due to gravity

    posted by Dr Russ
  6. The acceleration is
    g + (f/M) on the way up, and g - (f/M) on the way down. It is pointed down in both cases.

    They are different because both g and friction point down on the way up, but are in opposite directions on the way down. For that reason, it arrives at the ground at a lower velocity than it took off.

    posted by drwls

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