00:01
Okay, this tells us that a ball is kicked from the ground into the air.
00:11
So when the ball is at this height, so i'll call this height y1 of 12 .5 meters, then its velocity in the x direction, vx1 is going to be 5 .6.
00:43
Meters per second velocity in the y direction at point one is going to be um four point one meters per second so um y equals y initial which is zero plus v y initial which would be v initial, sine theta t, minus one -half g t squared.
01:39
So at y1, we've got vy1t, y1 t minus one half g t square so we know v y1 we know y1 we know v y1 so we need to figure out t here so i'm going to go ahead and graph it um y1 is 12 .5 or rather i'm going to subtract 12 point five off in the end.
02:41
So vy1 is going to be 4 .1 times t minus 9 .81 over 2 t squared.
03:00
And then i'm just going to subtract 12 .5 is y1.
03:07
And then i'm going to see when that equals zero.
03:14
So i'm going to write y equals maybe to make this thing plot.
03:19
Oh my goodness, it never equals zero.
03:24
How could that be? y1.
03:38
That's because this is not vy1.
03:45
This would be, well, v0 sine theta or let's write it as vy0.
04:06
This is at the beginning.
04:13
Okay, so let's write about vy1.
04:18
Vy1 equals the initial velocity minus gt.
04:29
So therefore, vy0 would be vy1 plus g t so i can substitute this into here so let me do that in desmos so vy1 plus gt now what do we have this crosses the x -axis at 1 .232.
05:36
So t equals 1 .232 seconds.
05:53
And what was i plotting on the y -axis? i was plotting y is a function of t.
06:24
So i wrote y equals, oh, but that's that's not really y equals.
06:31
Because in the end, i subtracted off.
06:41
That won't let me move over.
06:48
There we go.
06:50
So i plotted vy0 -0 -t minus 1 1⁄2t2t2t2 .5.
07:00
Okay.
07:02
The point is i was just looking for t equals zero.
07:05
And that's where i found.
07:06
That's what i found.
07:08
I just found where this equals 0, which is at 1 .232 seconds.
07:15
Okay, so what was i trying to figure out? so that was the time that it's going to reach y1.
07:31
Okay, so vy0 is going to be vy1.
07:40
So let's do that.
07:43
Vy1 and by the way i'm going to get rid of this and then i'm going to write t equals 1 .232.
07:56
That's all i wanted to figure out.
07:59
Okay, so t is 1 .30232.
08:04
Y0 equals vy1.
08:07
Vy1 is 4 .1.
08:09
What in the world is that? plus dt.
08:28
Right, times t, which is 1 .232.
08:34
Okay.
08:36
So, v initial, which i'm just going to call v here, is 16 .2.
08:58
Actually, since it's in the y direction, i'm going to call a w.
09:01
The x direction, i'll call v.
09:03
I'm going to call this w.
09:06
Okay.
09:06
Now, vx at 1 is 5 .6, but there's no acceleration in the x direction.
09:20
So the velocity for the whole time is going to be 5 .6 meters per second.
09:27
So i'm going to write velocity in the x direction is 5 .6 meters per second...