Aball is on the end of a rope that is 1.72 m in length. the ball and rope are attached to a pole and the entire apparatus, including the pole, rotates about the pole's symmetry axis. the rope makes an angle of 64.0° with respect to the vertical. what is the tangential speed of the ball

Tangential Speed equals 5.57m/s

Explanation:

In the figure shown for equilibrium along y- axis we have

Resolving Forces along y axis we have

Similarly along x axis

............(ii)" alt="\frac{v^{2} }{r}" />............(ii)" />

Dividing ii by i we have

In the figure below we have

Thus solving for v we have

Applying values we get

v=5.576m/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.