A 0.50 g oil droplet with charge +5.0 x 10-9 C is in a vertical퐸⃗-field. In what direction should the퐸⃗-fieldpoint and what magnitude should it have so that the droplet moves at constant speed? Ignore the forces that air exerts on the droplet.

The objects are acted upon by only one force. gravity. mass times the acceleration of gravity (-9.81meters/second squared).

250✖️100 to get you answer

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.