Answer
(a) The cliff is 255 meters high.
(b) If we ignore the time it takes the sound to reach us, then
the cliff is 314 meters high, which is an overestimate.
Work Step by Step
Let $h$ be the height of the cliff. Let $t_1$ be the time that the rocks falls to the bottom. Let $t_2$ be the time the sound travels up the cliff. Note that $t_2 = 8.00~s - t_1$.
$h = \frac{1}{2}gt_1^2 = (4.9~m/s^2)~t_1^2$
$h = (330~m/s)(t_2) = (330~m/s)(8.00~s-t_1)$
$h = 2640~m - (330~m/s)~t_1$
We can equate the two equations for $h$.
$(4.9~m/s^2)~t_1^2 = 2640~m - (330~m/s)~t_1$
$(4.9~m/s^2)~t_1^2 + (330~m/s)~t_1 - 2640~m = 0$
We can use the quadratic formula to find $t_1$.
$t_1 = \frac{-b\pm \sqrt{b^2-4ac}}{2a}$
$t_1 = \frac{-(330)\pm \sqrt{(330)^2-(4)(4.9)(-2640)}}{(2)(4.9)}$
$t_1 = 7.22~s, -74.6~s$
Since the negative value is unphysical, the solution is $t_1 = 7.22~s$.
$h = (4.9~m/s^2)(7.22~s)^2 = 255~m$
The cliff is 255 meters high.
(b) If we ignore the time it takes the sound to reach us.
$h = (4.90~m/s^2)(8.00~s)^2 = 314~m$
We would have overestimated because we would have supposed that the rock had been dropping for 8.00 seconds instead of just 7.22 seconds.
