At a point O on its path, a projectile has speed V and is travelling at an angle a above the horizontal. Derive the equation of the trajectory of the projectile in the form
y = x*tan(a) - gx2*(1+tan2(a))/(2V2)
Now, an aircraft is flown on the path given by y = 0.28x - (2.35*10-4) x2. By comparing the equation with the standard trajectory equation found above, find the speed and direction of motion of the aircraft at the point (0,0). Calculate the time of flight between the two points on the aircraft's path where y = 0.
If the equation is given numerically, doesn't this mean tan(a)=0.28 constantly? Which would mean that the motion is not projectile at all since the direction is constant ... what's going on here? How should I try to solve the problem?
As for the first part, I don't have a clue how to get there. Maybe there is a way to solve it with energy conservation? In my original attempt, I noted V*sin(a) = u*sin(theta) - gt where theta is the initial angle of projection and then tried to substitute into y = u*sin(theta)*t - 1/2gt2, where t was replaced in terms of x, but somehow ended up with
y = x*tan(a) + gx2*(1+tan2(a))/(2V2)
i.e. the wrong sign on the x2 term!