Speed = distance/time S=d/t
S = 4m/2s
S= 2 m/s
Velocity = speed and direction
V=2 m/s to the right

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.

M= 700 kgvalue of g( gravity) is constant = 9.8 m/s² or 10 m/s² w = mg = 700*9.8 = 6860n or 6.680*10³ n

Ibelieve the answer above is c but im not sure