An object with mass 50 kg moved in outer space. When it was at location <10, -26, -6> its speed was 4.5 m/s. A single constant force <210, 300, -190> N acted on the object while the object moved from location <10, -26, -6> m to location <15, -17, -9> m. Then a different single constant force <90, 200, 170> N acted on the object while the object moved from location <15, -17, -9> m to location <20, -24, -4> m. What is the speed of the object at this final location

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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.

The answer to the question is (a)