Originally Posted by

**skeeter** hopefully you have sketched a picture.

let h = height of the ball above the ground

x = horizontal distance from the ball's shadow to the point on the ground directly beneath it

using proportions formed by similar triangles ...

$\displaystyle \frac{20}{12+x} = \frac{h}{x}$

$\displaystyle 20x = h(12+x)$

$\displaystyle 20 \frac{dx}{dt} = h \frac{dx}{dt} + (12+x) \frac{dh}{dt}$

finally, note that $\displaystyle h = 20-5t^2$ (using a "nice" approximation $\displaystyle g = 10 \, m/s^2$)

or

$\displaystyle h = 20-4.9t^2$ (using the more accurate $\displaystyle g = 9.8 \, m/s^2$)