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A steel ball is dropped onto a hard floor from a height of 2.15 m and rebounds to a height of 2.09 m. (Assume that the positive direction is upward.)
(a) Calculate its velocity just before it strikes the floor.
-6.49 m/s
(b) Calculate its velocity just after it leaves the floor on its way back up.
(c) Calculate its acceleration during contact with the floor if that contact lasts 0.0800 ms.
(d) How much did the ball compress during its collision with the floor, assuming the floor is absolutely rigid?

  • physics - ,

    b is a little tricky.
    Notice the height an object gets depends on v^2 (1/2 mv^2=mgh)

    so the velocity just after rebound, is propotional the the sqrtrt of the rebound(hf/hi)

    b. v=+6.49*sqrt(2.09/2/15) (I did not check the answer in a

    c. Force*time=changemomentum
    mass*a*time=mass(answerinB-answerinA) solve for a.
    notice you will ADD the two answers, the signs on the second term come from -(-6.49)
    d I am not certain of what the instructor has in mind here.

  • physics - ,

    The compression is based on the displacement of the ball during its time interacting with the floor. The initial velocity is the velocity of the ball just before it hits the floor, and the final velocity is 0. The acceleration is the acceleration while the ball is interacting with floor > (answer b- answer a)/ time on the floor

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