00:01
Good day.
00:02
The topic is about uniformly accelerating motion.
00:05
The motion of an object that is moving at a concert acceleration rate can be described using the following kinematic formulas.
00:13
We have d is equal to vit plus 1 half 80 squared and a is equal to vf minus vi over t, where a is the acceleration in meters per second squared, vf is the final velocity in meters per second, vi is the initial velocity in meters per second, and this is the displacement in meters and time is the time or t is the time in seconds now let us consider the case when roger observes that water balloon passes his window and he notices he notices that after 0 .83 seconds the water balloon hits the ground below the his uh his window or where roger is situated is at the height of 15 meters from the ground we wish to find the speed at which the water balloon passes by the window of roger and that also we wish to find where exactly from from which floor is the balloon is initially launched consider that balloon is at this point is initially launched or is launched at rest or that simply dropped and that one that and that one floor of each of every floor of this dormitory has a high of 5 meters.
01:34
All right.
01:35
So to solve for this one, it is necessary that we assume that we, that air resistance place no effect on the motion of the balloon so that we can say that the balloon is in freefall.
01:53
And as such, its acceleration is constant, which is the acceleration due to gravity.
02:00
And we take that as negative 9 .8 meters per second square.
02:04
So negative because gravity points down.
02:12
So now we start by solving for the speed at which the balloon passes by the window of roger.
02:21
So we use the first chemoomatic equation above which is d equals vit plus one half 80 squared.
02:29
When the balloon makes or descends down to a distance of 15 meters, so that's negative 15, because downward it takes 0 .83 seconds for that to happen like descending like 15 meters takes 0 .83 seconds and a here is negative 9 .8 meters per second square so we wish to find the vi in this case the velocity of the balloon right where it is at the point at the point when it is 15 meters above the ground or that's where the window of roger is.
03:14
So substituting will give us negative 15 equals viii times 0 .83 plus 1ā2 times negative 9 .8 times 0 .83 squared.
03:31
So solving gives negative 15 equals 0 .8.
03:36
0 .83 vi minus 3 .37561.
03:42
Or this is equal to .83 vi equals negative 11 .62439.
03:53
Or that's vi equals negative 14 meters per second.
04:07
The negative sign again indicates that at this point when it passes the window of roger, it's heading downward.
04:15
And the speed at which it is going down is equal to 14 meters per second.
04:21
All right.
04:25
So next, let us solve for the floor from where the balloon is released from rest.
04:32
So again, we note that one floor is equivalent to 5 meters height...