Asked by GraceW1
A piece of gum comes lose from underneath an elevator that is moving upward at a speed of 6.30 m/s. The gum reaches the bottom of the elevator shaft in 3.00 s. With what speed does the gum hit the bottom of the shaft?
Answers
Answered by
Henry
V = Vo + g*t.
Vo = - 6.3 m/s.
g = +9.8 m/s.
t = 3 s.
V = ?
Vo = - 6.3 m/s.
g = +9.8 m/s.
t = 3 s.
V = ?
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