A man named Bungkas climbed a palm tree in 1970 and built himself a nest there. In 1994 he was still up there, and he had not left the tree for 24 years. Suppose Bungkas
asks a villager for a newspaper, which is thrown to him straight up with an initial speed
of 12.0 m/s. When Bungkas catches the newspaper from his nest, the newspaper's
velocity is 3.0 m/s, directed upward. From this information, find the height at which the
nest was built. Assume that the newspaper is thrown from a height of 1.50 m above the
ground.

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 answer would be d. hope this :

Idon't  know if i got it wrong but i picked ( a )