physics
posted by Anonymous .
Hakeem throws a 11.0 g ball straight down from a height of 2.0 m. The ball strikes the floor at a speed of 7.8 m/s. What was the initial speed of the ball? Assume that air resistance is negligible.

Vf^2 = Vo^2 + 2g*h,
Vo^2 = Vf^2  2g*h,
Vo^2 = (7.8)^2  19.6*2 = 21.64,
Vo = 4.65 m/s. 
because.

First of all, Henry cannot be right because the ball was "thrown" at the ground. His formula is only valid if it was free fall. However, the ball gained acceleration from the throw. You have to use energy formulas. I have a similar question on my worksheet, but I don't have a mass so I think the my problem is wrong. You should use energy problems in your case, but I'm too lazy to do all the work. Here are the formulas you need.
E=mgh
E=(1/2)mv^2
Also, remember conservation of energy.
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