Calculus
posted by Mike
A rocket follows a path given by y=x1/91x^3 where distances are in miles. If the horizontal velocity is given by vx=1.4x, find the direction of the velocity when the rocket hits the ground (assume level terrain) if time is in minutes.
So far I've only taken the derivative for y and don't know what to do afterwards
Vx=1.4x
Vy=1(3/91)x^2

Damon
Maybe you mean
y=x  (1/91) x^3
if so
y = 0 at ground, start and finish
0 = x  (1/91) x^3
x = 0 works of course
x^2 = 91
x = 9.54
Vx = 1.4 x = 13.4 miles/min at ground
what is Vy ?
dy/dt = dy/dx * dx/dt = [1(3/91)x^2 ] Vx = [ 1(3/91)9.54^2 ] 13.4 miles/min
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