0:00
Hide there.
00:01
So for this problem, we are told that a lead ball is dropped in a lake from a diving board that is 6 .48 meters above the water.
00:11
So we're going to call this height age 6 .48 meters.
00:19
And we are also told that it hits the water with a certain velocity and then sinks to the bottom with the same constant velocity.
00:30
It reaches the button at a time that is equal to 5 .5 .0 .0 seconds.
00:53
So with this information, what we need to do is for part a of this problem, is to determine how deep is the late.
01:06
So we're going to calculate the deep of this.
01:13
How deep the depth of this lake.
01:17
So in order to calculate this, we're going to let the height of the diving board be age, as we state in here.
01:26
And assuming that the positive wide direction and set the coordinate origin at the point where it was dropped, which is when we start the clock, thus ay equals to h designates the location where the ball strides the water.
01:44
Now let the depth of the late bd as we put in here and the total time for the ball to descend to be capital t.
01:53
This one we're going to call it capital t.
01:58
Well, we already put it in there.
02:01
And then with this, the speed of the ball as it reaches the surface using the, equations from motion.
02:10
We know that.
02:11
We know that and we're going to use the equation that is that is that the square of the final speed is equal to the initial speed square plus two times the acceleration due to gravity times the height.
02:26
In this case, we know that the ball is drop.
02:29
So the initial speed is zero.
02:31
So we will find that the speed that we call just simply as the speed b is equal to the square root of two times acceleration due to gravity.
02:40
Times the hide h.
02:42
And the time for the ball to fall from the board to the lake surface is what we are going to call the time one.
02:51
And this time is equal to the square root or two times the height divided by the acceleration due to gravity.
03:11
So with this, now the time it depends, now the time it expands descended in the lake, we're going to call that the time t2, and that is equal to the death of the lay divided by the final velocity just before it reaches, just after it toge the surface of the lake.
03:35
And then we can write that this is just simply d divided by the expression that we found for the velocity.
03:41
Let me just move this a little bit.
03:44
And then that velocity is equal to the square root of two times, acceleration due to gravity times the height h.
03:52
So with this, we will find that the total time is equal to the sum of these two times that we just have obtained.
04:02
So we will have that this is the square root of 2 times h divided by the acceleration due to 2 gravity.
04:07
And this plus the death divided by the square root of 2 times, the acceleration due to 2 times the height h.
04:13
So with this, if we want to solve for the death, we will obtain that this is equal to the total time times the square root of 2 times the acceleration due to gravity times the height, h, and this minus 2 times the height h.
04:45
So now we substitute all of these numerical values...