A waterskier skis over a ramp with a length of 15' and a height of 4' at a constant speed of 30 ft/second. How fast is she rising as she leaves the ramp?

Let x = the horizontal base of the ramp, y = the vertical wall, and z = the hypotenuse with theta as the angle in between x and z.

I see the following facts:

$\displaystyle tan \theta = \frac{4}{15}$

$\displaystyle \frac{dz}{dt} = 30 ft/s$

$\displaystyle z = \sqrt{241}$

Obviously, we are trying to solve for $\displaystyle \frac{dy}{dt}$

I tried setting the problem up like this:

$\displaystyle y = z sin\theta$

$\displaystyle \frac{dy}{dt} = \frac{dz}{dt} sin\theta$

$\displaystyle \frac{dy}{dt} = 30(0.2606)$

$\displaystyle \frac{dy}{dt} = 7.818 ft/sec$

I don't think this is the right answer. Does anybody see where I am going wrong?