Answer
$1.2~s$ before it reaches the ground, the rock is $~~34.1~meters~~$ above the ground.
Work Step by Step
We can find the time it takes the rock to fall $60~m$:
$y = \frac{1}{2}at^2$
$t = \sqrt{\frac{2y}{a}}$
$t = \sqrt{\frac{(2)(60~m)}{9.8~m/s^2}}$
$t = 3.5~s$
Note that: $~~3.5~s-1.2~s = 2.3~s$
We can find the distance the rock falls in a time of $2.3~s$:
$y = \frac{1}{2}at^2$
$y = \frac{1}{2}(9.8~m/s^2)(2.3~s)^2$
$y = 25.9$
We can find the height above the ground $1.2~s$ before the rock reaches the ground:
$60~m - 25.9~m = 34.1~m$
$1.2~s$ before it reaches the ground, the rock is $~~34.1~meters~~$ above the ground.
