Asked by Vova
A person on a trampoline bounces straight upward with an initial speed of 4.7m/s . What is the magnitude of the person's speed when she returns to her initial height?
Answers
Answered by
Henry
d = (Vf^2-Vo^2) / 2g,
d = (0-(4.7)^2) / 19.6 = 1.13m,up.
Vf^2 = Vo^2 + 2gd,
Vf^2 = 0 + 2*9.8*1.13 = 22.15,
Vf = 4.7m/s.
d = (0-(4.7)^2) / 19.6 = 1.13m,up.
Vf^2 = Vo^2 + 2gd,
Vf^2 = 0 + 2*9.8*1.13 = 22.15,
Vf = 4.7m/s.
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