Hello, PFX!

A stone is projected from a point $\displaystyle O$ on a cliff with a speed of 20 m/s

at an angle of elevation of 30°.

t seconds later the angle of depression of the stone from $\displaystyle O$ is 45°.

Find the value of $\displaystyle t$.

The parametric equations for the position of the stone (relative to $\displaystyle O$) are:

. . $\displaystyle \begin{array}{ccccccc}x &=& (20\cos30^o)t & \Rightarrow & x &=& 10\sqrt{3} t \\

y &=& (20\sin30^o)t - 4.9t^2 & \Rightarrow & y &=& 10t - 4.9t^2 \end{array}$

Consider the position of the stone at time $\displaystyle t.$ Code:

*
* *
O* - - - +A
\ *:
\ :
\ :
*B

Since $\displaystyle \angle AOB = 45^o,\:\Delta OAB$ is an isosceles right triangle.

. . Hence: .$\displaystyle OA = AB$

This means: .$\displaystyle x \,=\,-y\quad\Rightarrow\quad (10\sqrt{3})t \;=\;-\left[10t - 4.9t^2\right]$

. . which simplfies to: .$\displaystyle 4.9t^2 - 10(1+\sqrt{3})t \:=\:0 $

. . which factors: .$\displaystyle t\left[4.9t - 10(1+\sqrt{3})\right] \:=\:0$

. . and has roots: .$\displaystyle t = 0,\;\frac{10(1+\sqrt{3})}{4.9} $

Therefore: .$\displaystyle t \:\approx\:5.6$ seconds.