Asked by Ariel
A stone is thrown upward a 20 m/s. It is caught on its way down 5.0 m above where it was thrown. How fast was it going when it was caught?
Answers
Answered by
Henry
h max = (V^2-Vo^2)/2g = (0-(20)^2/-19.6 = 20.41 m.
V^2 = Vo^2 + 2g*h = 0 + 19.6*(20.41-5) =
302.
V = 17.4 m/s
V^2 = Vo^2 + 2g*h = 0 + 19.6*(20.41-5) =
302.
V = 17.4 m/s
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