1. The problem statement, all variables and given/known data A girl throws a ball vertically upwards with speed 8ms^-1 from a window which is 6m above horizontal ground. (a) Find the greatest height above the ground reached by the ball. (COMPLETE) (b) Find the distance below the window of the point where the balls meet. http://img505.imageshack.us/img505/6465/55061996xe3.png [Broken] 2. Relevant equations Vertical motion under gravity (ignoring air resistance)- Releasing objects v=gt v^2=2gh h=1/2gt^2 Vertical motion under gravity (ignoring air resistance)- Projecting objects v=u-gt v^2=u^2-2gh h=ut-1/2gt^2 Where: v=final velocity u=initial velocity t=time g=acceleration due to gravity (count as 9.8ms^-2) h=height from ground(a.k.a. distance) 3. The attempt at a solution Well I have done the first part... (a) 2gh=u^2-v^2 h=(u^2-v^2)/2g =(8^2-0)/2*9.8 =3.27m 6+3.27=9.27m (FINAL ANSWER) However for (b) I do not know where to start. I know the answer is 0.104m (checked the answers) but I have no idea how to arrive there, I have simply done scribbles all over the place: Attempt 1 http://img505.imageshack.us/img505/8835/attempt1kv2.png [Broken] Attempt 2 (Using previous info) http://img508.imageshack.us/img508/9710/attempt2sw1.png [Broken] I hope someone can tell me where I am going wrong or give me a few tips on what to do and I should be able to calculate the rest. P.S. Thanks in advance! I'm not sure if this is the right section as this unit is part of the maths AS course but mechanics is mainly physics so I placed the question here - my apologies if it is in the wrong section.