A girl on top of a building drops a baseball from rest.at the same moment a boy below throws a golfball upwards towards her with a speed of 20m/s.the golfball is thrown from a point 18m below where the baseball is released.how far will the baseball have dropped when it passes the golfball.

Question

A girl on top of a building drops a baseball from rest.at the same moment a boy below throws a golfball upwards towards her with a speed of 20m/s.the golfball is thrown from a  point 18m below where the baseball is released.how far will the baseball have dropped when it passes the golfball.

Henry · Answer

d1 + d2 = 18 m. 0.5g*t^2 + Vo*t + 0.5g*t^2 = 18 4.9t^2 + Vo*t - 4.9t^2 = 18 Vo*t = 18 20t = 18 t = 0.9 s. d1 = 0.5g*t^2 = 4.9 * 0.9^2 = 3.97 m.