Answer
(a) The time to reach the bottom of the cliff is 5.80 seconds.
(b) The speed just before hitting the bottom is 41.3 m/s.
(c) The total distance the stone traveled is 99.6 meters.
Work Step by Step
(a) $y = y_0 + v_0t + \frac{1}{2}at^2$
$-75.0 = 0 + (15.5)t -4.90t^2$
$4.90 t^2 -15.5~t - 75.0 = 0$
We can use the quadratic formula to solve for t.
$t = \frac{-b \pm \sqrt{b^2-4ac}}{2a}$
$t = \frac{15.5 \pm \sqrt{(-15.5)^2 - (4)(4.90)(-75.0)}}{(2)(4.90)}$
$t = -2.64 s, 5.80 s$
The negative value for t is unphysical, so the time to reach the bottom of the cliff is 5.80 seconds.
(b) $v = v_0 + at = 15.5 ~m/s - (9.80 ~m/s^2)(5.80 ~s)$
$v = -41.3 ~m/s$
The speed just before hitting the bottom is 41.3 m/s.
(c) Let $y_{max}$ be the maximum height above the top of the cliff.
$y_{max} = \frac{v^2-v_0^2}{2a} = \frac{0 - (15.5 ~m/s)^2}{(2)(-9.80 ~m/s^2)} = 12.3 ~m$
Then the total distance is 12.3 m + 12.3 m + 75.0 m = 99.6 meters.
