00:01
In this question here, we are given that a rock is dropped from a 200 feet building.
00:07
So we have a building.
00:12
So we are dropping a rock from here.
00:17
This building is 200 feet.
00:21
So this distance is 200 feet.
00:25
The position of the rock in feet above the ground is given by s of t equals negative 16 t square.
00:36
Plus 200 but t is the time in seconds since it was dropped so the first part of the problem is asking us to find what is the velocity one second after being dropped we know that velocity that v of t refer to the velocity that's derivative of the position so that is negative 16 derivative of t squared is 2t derivative of 200 is 0 so that's negative 32 t that's my velocity we want to find the velocity one second after being dropped so we want to calculate what is v of 1 so that's negative 32 times 1 which is negative 32 feet per second the velocity is negative meaning it's going in the negative wide direction now the second part says what's the velocity 2 seconds after being dropped so v of 2 is negative 32 times 2 which is negative 64 feet per second.
01:45
Again, the velocity is negative denoting that it's traveling in the negative y direction.
01:52
The second part of the problem asks, when will it hit the ground? now s of t refers to its distance from the ground t seconds after being dropped.
02:05
Now it will hit the ground when the distance between the rock and the ground is zero...