A rocket follows the path $(y-40)^2=160x$ after traveling vertically to the height of 40 feet (making for a very convenient continuity.
The problem asks "If the component velocity in the vertical direction is constant at $v_y=180$, determine the magnitude of the components of the rocket's velocity and acceleration when it reaches an altitude of 80 meters."
Also, $v_y=180$.
I have made the following approaches to solve the problem:
Expand to:
$$x=(y^2-80y-1600)\frac{1}{160}\\\dot{x}=\frac{1}{160}(2y\dot{y}-80\dot{y})\\\dot{y}=180$$
Where we simply subsitute for $y$ and $\dot{y}$. However, while I got it to yield the same result, the answer is 201 whereas my result is 90 for both cases!
$$y=\sqrt{160x}+40\\\frac{d}{dx}f(x)=\frac{d}{dx}g(x)\space\mathrm{ IF }\space f(x)=g(x)\\
\frac{d}{dt}(\sqrt{160x}+40)=180\\\dot{x}=\frac{360\sqrt{x}}{\sqrt{160}}$$/
I would be grateful to anyone who could tell me what I'm doing wrong.
Once you've determined the $x$ and $y$ components of the velocity, namely $v_x = \dot{x} = 90$ and $v_y = \dot{y} = 180$, to get the magnitude of the velocity, use the Pythagorean Theorem: $v = \sqrt{v_x^2+v_y^2} = \sqrt{90^2+180^2} \approx 201.24$.