While on vacation, Paul is staying in a hotel room with a balcony 18.0 m above the ground that is overlooking the ocean. He is curious as to how far he could throw a
rock before it hits the water. Recognizing that it would be unsafe to actually throw
the rock, he decides to work out the answer mathematically. He knows that he can
throw the rock horizontally at a speed of 30.0 m/s. How far from the hotel will the
rock be when it hits the water?

Cpowerlifting my boi

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.