00:01
And in this problem here, we have a water jet, which is firing from the ground here in a fountain.
00:08
Let's say, call that zero, which is zero feet.
00:14
And the water launches up into the air up to xm for max of 240 feet.
00:23
And we have a few questions here regarding this water.
00:27
And what we do know is that we only have one force acting on the water.
00:31
And that is gravity, which is, we'll say that the force is minus g, because gravity acts downwards.
00:42
So we have a, let's say, x, x that goes like this.
00:50
And we have three questions here.
00:54
We have a.
00:56
What is the initial velocity of the water? b, what is the velocity when it's falling down? down at half of its maximum value and see later than what is the time delta t to fall from x and half on the way down to zero.
01:23
So we'll start with a like this and we have some boundary conditions which need to be satisfied.
01:36
We do know that when we go from the bottom up to the top, we have initially, we're initially at x0 equals zero.
01:52
We're going to reach a maximum height of 240 feet.
01:58
We start with the velocity v nod, which we don't know.
02:04
We find it and we finish the velocity of.
02:08
Of zero.
02:10
So on the top of our fountain here, we have zero velocity.
02:19
And we're going to use newton's laws to find these values.
02:24
And as you know, we have f equals an a, but you can also just write this.
02:37
We know that we have a equals d2.
02:45
X d t squared and from integrating this expression we will find a few different equations the first one is we can say is v equals a t plus v zero and the second one is x equals if you integrate the above equation one more time a t squared over 2 plus b0 t plus x not.
03:25
And important note here, because we only have one acceleration which is acting on our water, we have a equals minus g.
03:39
Okay, so we're going to use these equations...