00:01
Okay, so the question wanted us to find out how fast are the balloons traveling when they pass the rogers window? also, want us to find out where are the balloons from? like where were the balloons being released? well, let's find out.
00:17
Well, we final square minus v -i -square over 2g, secret.
00:22
We find that is the final speed, which is when the speed, when the balloons hit the ground, and vi, we consider it.
00:30
Is the speed that passed by rogers window because we only consider the motion from rogers window to the ground.
00:38
So the initial speed should be the speed that passed by rogers windows, okay? g is the acceleration of the gravity.
00:46
H is the height from the ground to the rogers window, which is the third floor of the door, which is 15.
00:52
Also, we know v -final can be equal to v -i plus gt.
00:55
Since there's initial speed, so it's initial speed plus, acceleration of gravity times time.
01:03
And then we have v -i -plus g -t as we final.
01:06
If we're plugging to the v -final square minus v -i -square, which is v -i -plus g -t to a power square minus v -i -square over 2 -ch equals h.
01:14
And then we have 2 -vi plus g -t times gt over 2g -2 -h.
01:19
Okay? and then, so how can i get 2v -i plus gttt times gtt? you know, remember, do you guys remember, a -square minus b -square, is equal to a plus b a minus b and this formula works here as well.
01:36
Then we'll have, let me see, let me clear this.
01:42
Oops, then we have 2vi, sorry about it, so we have 2v times gt plus g square t squared is equal to 2gh...