Starting from rest, an ice skater of mass 54.0 kg pushes away from the boards with a force of 130.0 N [W] and moves 0.704 m. He then moves at a constant velocity for 4.00 s before he digs in his skates and causes a net force of 38.0 N [W] to slow him down until he stops. How far, in total, did he move?

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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.

D. the index of refraction of the second medium is lower.