a simple pendulum has a length of 52.1 cm and makes 83.7 complete oscillations in 2.00 min.

(a) find the period of the pendulum.

(b) find the value of g at the location of the pendulum.

Time period is 120/83.7=1.43 s and
g = l / [(T / 2pi)^2] =0.521/[(1.43/2pi)^2]=10 m/s

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 actual answer is c force stays the same while direction changes