Post a New Question


posted by .

The fist ball is thrown up ward with avelocity of 20m/s at the same instant. The second ball is dropped at 40m hight from the first ball at what time the two ball collide each other?

  • physics -

    They collide when Y1 = Y2. Those are the heights of the two balls, as measured from the a defined "zero" height location.

    Y1 = 20 t - 4.9 t^2
    Y2 = 40 - 4.9 t^2

    When Y1 = Y2,
    20 t = 40
    t = 2 seconds
    Y = 40 - 4.9 t^2 = 20.4 m above the ground

Answer This Question

First Name:
School Subject:

Related Questions

More Related Questions

Post a New Question