A circus acrobat is shot out of a cannon with an initial upward speed of 34 ft/s. if the acrobat leaves the cannon 4 ft above the ground, how long will it take him to reach a net that is 8 ft above the ground

Distance covered, S = 8ft - 4ft = 4 ft

Initial velocity, u = 34 ft/s

The acrobat is moving upwards and therefore, a = - g = - 9.81 m/s^2 = 32.19 ft/s^2

Applying equation of motion;

S = ut - 1/2gt^2

Then,

4 = 34t - 1/2*32.19t^2

4 = 34t - 16.095t^2

Rearranging;

-16.095t^2 + 34t - 4 = 0

Solving the quadratic equation;

t = {-34 + / - Sqrt [34^2 - 4 (-16.095) (-4) ]}/2 (-16.095) = 1.06 + / - 0.93 = 0.13 or 1.99

The 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.