1. The problem statement, all variables and given/known data a ball is thrown in angle α from height H , moving horizontal distance D. express the highest spot it reaches using the parameter D. 2. Relevant equations r=r0+v0(t)t+0.5gt2 3. The attempt at a solution My question is why I can;t simply find the highest spot using the vectors without using the equation? The velocity vector is integrated from the acceleration so it already contains the acceleration so I dun get why I get a wrong answer. Here is what I did: a=(0,-g) v=∫a +v0=> v=(vcos(α),vsin(α)-gt) now I found t by saying highest spot is when vy=0 so I get vsin(α)-gt=0 => t=( vsin(α) )/g Now here is what I am supposedly doing wrong: The position vector in any given time is: ∫v+r0 => r(t)=(-vsint(α)t,H+vcos(α)t-gt2) Ok now why can't I simpy calculate it as r(t=( vsin(α) )/g) ? Why I have to use the equation r=r0+v0(t)t+0.5gt2? I am looking for the D it will reach in the given time (which I calculated as the time it takes the ball to reach it's highest point). The acceleration is already in account so what am I missing? the next steps I'd take are saying: rx=D and finding t using D. ry=0. using the t I just found (all this to get D involved). (here again I would just place t in teh vector but the answer actually contains the equation of r=r0+v0(t)t+0.5gt2 ) why can;t I use it without it? And eventually using the new ts I just found (which contain the D I need) Place it again in the vector which expresses highest height. I know in a specific t it is in a specific position , I wanna find the t it reaches with specific speed....why do I need the equation if I have specific points? There is no problem using for instance the place it reaches in t=3. I dun get it.