Asked by Anonymous

When a ball is thrown upward, it experiences a downward accelera-
tion of magnitude 9.8 m/s
2
, neglecting air resistance. With what
velocity must a ball leave a thrower’s hand in order to climb for 2.2 s
before stopping?

Answers

Answered by Henry
V = Vo + g*t
V = 0
Vo = Initial velocity.
g = -9.8 m/s^2
t = 2.2 s.
Solve for Vo.


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