Recent content by moria

You are right, when I do that I get 1.06 seconds and when I add it to 1.96 I get 3.02 seconds. And doing this gets me the same result, 3.02 seconds. Thanks for the help.

With the risk that I might write bullsh*t: 25+d=vx*t d=vx*t-25 tg(37)=h/vx*t-25 h=tg(37)*(vx*t-25) h=vy*t-4.05t² tg(37)*(vx*t-25)=vy*t-4.905t² Like this?

Vy=sin(55)*v and Vx=cos(55)*v v is 22.22 m/s What to do with the 25 meters? I know how to solve it when the projectile is fried from the bottom of the ramp. tg(alpha)=Sy/Sx tg(alpha)=(vy*t-gt²/2)/(vx*t) and so on, but where do the 25m fit into all that?

Shouldn't the path of of the trajectory be the same as when I'm solving it without slope? Wouldn't solving the height through tg(37)*14.9 change the path? Clear me up if I'm wrong, cause that is the part that bothers me the most about this problem.

Homework Statement My problem looks like this: http://i.imgur.com/3M9wE43.png I need to find the point of impact and the time the body traveled. Homework Equations The Attempt at a Solution I tried to find the time for 25 meters and with that the vertical velocity and height. Ended up...