This is from Applications of the Derivative under Curvilinear Motion:
So a rocket follows a path given by y=x-(1/90)x^3. Distance in km.
If the horizontal velocity is Vx=x, find the magnitude and direction of the velocity when the rocket hits the ground, if time is in minutes.
So I have so far = I do the product rule
dx/dy=1-(1/90)x^3 + x-(1/30)x^2(dx/dt)
So then I multiply by 90 to get rid of the 90 and 30.
So I have now
Gives me now
Vy = -1.1x^3+90x
I am stuck at this part and I don't know if I did this correct??
How do I solve with Vx=x???
Can someone direct me in the right direction.