1. The problem statement, all variables and given/known data An elastic ball is projected with speed V from a point O of a smooth plane inclined at an angle theta to the horizontal. The initial direction of projection is up the plane, makes an angle theta + phi with horizontal and lies in a vertical plane containing a line of greatest slope of the plane. (i)Calculate the distance between O and the point of the first bounce. (ii)If the ball ceases to bounce at the instant it returns to O show that tan(theta)tan(phi)=1-e 2. Relevant equations The way I tried this for both parts was to make the x axis follow the line of the slope NOT the horizontal which means splitting gravity into components, but this became really confusing. 3. The attempt at a solution I chose the x axis to be the line of slope and z to be at right angles to this as the other axis. Using F=ma and F=mg I found a=g and the integrated to find the position vector r(t) = g(t^2/2)+vo, where vo is the starting velocity in component form, i.e. vo = kVsin(phi) + iVcos(phi), where k and i are unit vectors. Then because gravity has to be split up I used vector g = -igsin(theta)-kgcos(theta). Plugging both vector g and vo into the r(t) equation and splitting again into components, I get z(t) = Vtsin(phi)-(t^2/2)gcos(theta) and x(t)=Vtcos(phi)-g(t^2/2)sin(theta). When the ball hits the slope z(t) = 0 so using my equation for z(t) and rearranging to get t, I find t=2Vsin(phi)/gcos(theta). Then I put this t into x(t) and that should be the distance up the slope. To find the distance from O along the horizontal I used trig, but it comes out with this horrible looking equation and I'm pretty sure my error is in the gravity stuff because the change in axis confuses me massively! Sorry this is really rambling, I hope it makes sense.