Answer
52 meters.
Work Step by Step
Let down be the positive direction and y = 0 be the initial height. The initial velocity is 0, and the acceleration is $g = 9.8\; m/s^2$. The final position is the unknown, y = H. The time of fall is $t_1$.
Apply equation 2–11b.
$$H=0+0+\frac{1}{2}gt_1^2$$
For the sound wave, we know that $v_{sound}=\frac{H}{3.4s-t_1}$. The time of travel for the sound wave is the total travel times of 3.4 seconds, minus the drop time.
This can be re-arranged to solve for $t_1=3.4s-\frac{H}{v_{sound}}$.
Substitute this expression into the earlier equation and solve for H.
$$H=0+0+\frac{1}{2}g(3.4s-\frac{H}{v_{sound}})^2$$
This results in a quadratic equation for H.
$$\frac{g}{2v_s^2}H^2-(\frac{g(3.4s)}{v_{sound}}+1)H+\frac{1}{2}g(3.4s)^2=0$$
Putting in numbers, we obtain H = 52m and H = 25900 m. The second solution is not correct because it gives a negative time for the rock to fall.
The cliff is 52 meters high.
