Question

A construction worker ascending at 6m/s in an open elevator 42 m above the ground drops a hammer.

Answer 1

Let us follow the motion of the hammer first. Because the elevator is in motion, when he drops the hammer, because of inertia, there is a slight moment when the hammer also rises with the elevator. Eventually it will reach its highest peak and drop down to the floor. So, the total time for the hammer to reach the floor would include: (1) the time for it to rise with the elevator to its highest peak and (2) the time for the free fall from the highest peak to the floor.

1.) Time for it to rise with the elevator to its highest peak:
      Hmax = v²/2g = (6 m/s)²/2(9.81 m/s²) = 1.835 m
      Time to reach 1.835 m = 1.835 m * 1 s/6 m = 0.306 s
     
     Time for the free fall from the highest peak to the floor:
      t = √2y/g, where y is the total height
      y = 1.835 m + 42 m = 43.835 m
      So,
      t = √2(43.835 m )/(9.81 m/s²) = 2.989 s

      Therefore, the total time is 0.306 s + 2.989 s = 3.3 seconds

2.) Velocity of impact of a free-falling body is:
      v = √2gy
      v = √2(9.81 m/s²)(43.835 m)
      v = 29.33 m/s