The question:

Boy saw a ball first moving up and then down past an open window. The top-to-bottom height of window is 2.00m. Ball in view for a total duration of 0.50s. Calculate how much further above the window top did the ball move before reaching its highest point. [acceleration due to gravity = 10ms^-2 and ignore air resistance].

 

Answer is given, but i am unable to solve it. Tried using 1/2mv^2 and mgh etc., but still unable to. Anyone out there who knows...?

You might want to try to use the equation(s) of uniform acceleration?

 

Make sure to post here again if you want more help. My answer is short because it is my intention to let you do most of the work (that's our policy here in the homework section).

Part of this is a matter of how much information is actually contained in the setup of the problem. For example, you are told the ball is visible for 0.5 sec. What will the time be for the ball moving up as compared to moving down?