Asked by Charles
A ball is dropped from a roof to the ground 4.00 m below. Exactly 0.400 s later, a rock is thrown downwards with some initial velocity. If the two objects hit the ground at exactly the same time, what was the initial speed of the rock?
Answers
Answered by
Henry
d = Vo*t + 0.5g*t^2 = 4m
0 + 4.9t^2 = 4,
t^2 = 0.8163,
t1 = 0.90s.
t2 = 0.9 - 0.4 = 0.50s.
d = Vo*t + 0.5g*t^3 = 4,
0.5Vo + 4.9*(0.5)^2 = 4,
0.5V0 + 1.225 = 4,
0.5Vo = 2.775,
Vo = 5.55m/s.
0 + 4.9t^2 = 4,
t^2 = 0.8163,
t1 = 0.90s.
t2 = 0.9 - 0.4 = 0.50s.
d = Vo*t + 0.5g*t^3 = 4,
0.5Vo + 4.9*(0.5)^2 = 4,
0.5V0 + 1.225 = 4,
0.5Vo = 2.775,
Vo = 5.55m/s.
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