Aproton traveling at 27.1° with respect to the direction of a magnetic field of strength 2.33 mt experiences a magnetic force of 6.54 × 10-17 n. calculate (a) the proton's speed and (b) its kinetic energy in electron-volts.

(a) The magnetic force experienced by a charged particle is:

where
q is the charge
v is the velocity of the particle
B is the magnitude of the magnetic field
is the angle between the directions of v and B

The proton in our problem has a charge of , and it travels through a magnetic field with strength 

The direction between its velocity and B is , and the force exerted on the proton is . Re-arranging the previous equation and using these data, we can find the value of v:


(b) Using the mass of the proton, , we find its kinetic energy:


And keeping in mind that , we can convert this value into electronvolts:

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.

Ibelieve the answer above is c but im not sure