Answer
(a) The cliff is 532 meters tall.
(b) The teacher will be moving at a speed of 102 m/s
Work Step by Step
(a) We can find the distance $y_1$ the teacher falls in 3.0 seconds.
$y_1 = \frac{1}{2}gt^2 = \frac{1}{2}(9.80~m/s^2)(3.0~s)^2$
$y_1 = 44.1~m$
Let $h$ be the height of the cliff. The sound traveled a total of $2h - y_1$ meters in 3.0 seconds.
$v = \frac{2h-y_1}{t}$
$h = \frac{vt+y_1}{2} = \frac{(340~m/s)(3.0~s)+44.1~m}{2}$
$h = 532~m$
The cliff is 532 meters tall.
(b) $v^2 = v_0^2+2gh = 0+2gh$
$v = \sqrt{(2)(9.80~m/s^2)(532~m)}$
$v = 102~m/s$
The teacher will be moving at a speed of 102 m/s
