00:01
In this problem we have been given that there is a ball that's dropped in the well and the sound of the splash is heard five seconds later.
00:12
So here the speed of the sound, if it is considered as 3 .30 meter per second, we need to figure out the depth of this well.
00:21
So let's consider the depth of the well as d which we need to compute here.
00:25
And as the sound is heard five seconds later.
00:31
That means, so here if we compute, if we relate the distance, let's say the sound took t seconds to reach up.
00:42
So the stone took 5 minus t seconds to come down or to fall down.
00:48
So here we can frame the equation because the sound moves with the speed of 3 .30 meter per second and the distance traveled is t in time t.
00:56
So we can say that d will be equal to 3 .3.
00:59
30 times t so that's equation one and now the stone falls so we can use the equation s equals ut plus half a t square and we can relate the displacement that is d initial velocity with which the stone falls is zero so that will be 4 .9 5 minus t whole square because now the time taken will be 5 minus t so from here we have to figure out d so we can replace d here so we can replace d here so so from equation 1, we can get the value of d.
01:34
So that would be d over 330.
01:38
So we can put this value here, we get d is equal to 4 .9 into 5 minus d by 330 whole square...