00:01
In this problem, we have been given that a brick is dropped from the top of the building.
00:06
So its initial velocity will be 0 meter per second.
00:09
And after t is equal to 2 .5 seconds.
00:14
So after this time interval, the brick hits the surface, that's the ground.
00:19
So we need to determine the height of this building.
00:22
So it's better to figure out just the displacement using the quantumatic equation ut plus half a t square equal to s.
00:29
And we know that that acceleration here will be minus 9 .8 meter per second square and the negative sign indicates that the direction of acceleration is in the downward direction.
00:41
So here we will see when we put the values, we're going to get s as 4 .9 times the square of 2 .5.
00:50
That is minus 30 .6 to 5 meters.
00:54
So the displacement here is minus 6.
00:58
Minus 30 .6 to 5 meters.
01:00
And the height here will be 30 .625 meters.
01:05
So we can see that the building is 30 .6 to 5 meters tall approximately...