Asked by olawale
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.
Answers
Answered by
Henry
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.
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.
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