physics

A ball is thrown vertically upward with a speed of +12.0 m/s.

How long does the ball take to hit the ground after it reaches its highest point?

What is its velocity when it returns to the level from which it started?

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1. Well, the easy way is to say the average speed of the ball on the awy up is (12 + 0) / 2 = 6 m/s

Now how high did it go?
(1/2) m v^2 = m g h
144/2 = 9.8 h
h = 7.35 meters up
so
6 t = 7.35
t = 1.22 seconds

part 2 is trivial, same speed down as up but opposite direction, -12 m/s

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posted by Damon

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