Hi, I'm stuck in my maths assignment and need help with one of the questions. “A stunt motorcyclist launches himself from a ramp inclined at 30 degrees to the horizontal. He aims to clear a line of cars that extends to a distance of 40 metres from the end of the ramp. Use calculus methods to determine the minimum possible take-off speed for the rider given that the end of the ramp is 3 metres above the ground and an average car is about 1.5 metres tall.” OK, so my teacher told me to treat the model as a coordinates on a Cartesian plane or something like that. At the bottom right of the ramp I put the origin there so i don't have to deal with the negative vertical axis. I used i for the horizontal axis and j for the vertical axis. I have been given the equation v=v*√3/2 i+v/2 j from the ramp using sine and cosine, but I'm not sure if thats to work out the velocity at the point where the motorcycle leaves the ramp. Now, I'm going to have a wild guess and say if I differentiate velocity I will get the displacement. But the problem with that is that it adds another variable, t, also i need to find c when i differentiate. I need help finding the height (j) for the midpoint where the vertical velocity is zero. Since the question asks for the minimum velocity I was thinking that the midpoint would be at 20m or 20i. The path trajectory also needs to hit 40i+1.5j. So my question is; 1.) How to find the constant (c) when differentiating velocity. 2.) How to find the height of the midpoint for the flight path. If anyone could just point me in the right direction it would be very helpful. Also I have included two diagrams as attachments which I drew for the question.