00:01
A rock is dropped from the top of a 60 meter tall building.
00:05
So let's say that y sub -zero is 60 meters.
00:14
How far above the ground is the rock one in two seconds before it reaches the ground? okay, well, y is going to equal y initial, those v initial t, which is zero minus one -half gt squared.
00:43
Now, to reach the ground, that means that this would be zero equals y -0 minus one -half g -t to the ground squared.
00:57
So y -0 times two over g, and then take the square root of that is going to be time to the ground.
01:14
But we want a time that is 1 .2 seconds before that.
01:21
So that would be the square root of 2y0 over g minus t.
01:34
So how far above the ground will it be at that time? well, let's figure out what this time would be.
01:45
I'm going to call this t sub 1.
01:48
So putting that in a calculator with y sub 0 equals 60 and t equals 1.
02:00
Then t sub 1 is going to equal the square root of 2 y sub 0 over...