Posted by **Jake** on Sunday, September 27, 2009 at 12:21pm.

A ball is thrown vertically into the air and when it returns after an interval of 2 seconds it is caught. What is the acceleration of the ball after leaving the hand?

I have no idea how to solve this!

- physics -
**drwls**, Sunday, September 27, 2009 at 1:03pm
You have not been given enough information to answer this problem.

In general, for objects falling near the surface of the earth, the acceleration is -g = -9.8 m/s^2, no matter how long it takes to come down. The minus sign indicates downward acceleration

If you did not know the value of g (for example, if you were on an unknown planet), you could determine the acceleration (a) if you knew the launch velocity of the ball (V). The time it takes to come down (T) is given by

a T/2 = V

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