if a 3.0-kg mass is now attached to the end of the spring, and the mass is allowed to fall, where will the end of the spring line up with the ruler marks when the mass is at its lowest position?

Hypothesis. why don’t you know this? i learned this stuff in like 5th grade

Distance covered, s = 8ft - 4ft = 4 ftinitial velocity, u = 34 ft/sthe acrobat is moving upwards and therefore, a = -g = -9.81 m/s^2 = 32.19 ft/s^2applying equation of motion; s = ut -1/2gt^2then,4 = 34t -1/2*32.19t^24 = 34t - 16.095t^2rearranging; -16.095t^2 + 34t - 4 = 0solving the quadratic equation; t = {-34 +/- sqrt [34^2 - 4(-16.)]}/2(-16.095) = 1.06 +/- 0.93 = 0.13 or 1.99the acrobat will take 0.13 seconds to be at 8 ft above the ground. additionally, once at the maximum height, the acrobat will move down and be at the same spot (that is height) after 1.99 seconds moving down.

F= (mass)(acceleration) = mam = 0.25 kgvi = 16 m/st = 2 svf = 0 m/s (since it was put to stop)a=(vf-vi)/ta=(0-16)/2a = 8 m/s^2 (decelerating)f = ma = (0.25 kg)(8 m/s^2)f = 2 nhope this !