00:01
All right, there is a figure in the book, and it says that a tennis ball, first of all, is being dropped.
00:11
So for the tennis ball, the height initially of the tennis ball is 54 meters, and it's just being dropped.
00:34
And so, y, as a function of time, is going to be, and you know what, i'm going to write y at the collision is going to be y of the tennis ball initial minus v0, which is zero, or plus v0, which is zero.
01:07
There's no initial velocity for the tennis ball because it's just being dropped.
01:10
T minus one half g t squared, where g is 9 .81 meters per second squared.
01:23
That's gravity.
01:25
Okay.
01:26
So now, and by the way, the tennis ball is always at x collision, which is 50 meters.
01:48
Now, let's think about the rock.
01:53
The rock started at x equals 0.
01:58
So let's figure out, let's write an equation for x for the rock.
02:05
X of collision is going to be x0, which is 0, plus vx0.
02:19
That's what we're trying to figure out.
02:23
So that's velocity of the rock, v.
02:34
Are initial in the x direction.
02:42
To put it in the x direction, we need to multiply by the cosine of theta.
02:53
X equals x initial plus v initial, t, and i think it's telling us that the rock and the ball were launched or dropped at the same time, at the same instant.
03:22
Yes.
03:23
So x equals x initial plus v initial, t and then there's no acceleration in the x direction now y for the rock at collision which would just be y collision would be y initial which is zero plus v0 t which would be v0 t which would be v of the rock initial sine theta to put it in the y direction t minus one half g t squared okay we know that t equals three and we know that y equals 10 why collision is 10 and we know that t of the collision is three i'm thinking they're giving us too much too much information here because looking at the first equation.
04:57
Y collision we know.
04:59
Y of t0 we know.
05:03
G we know.
05:06
And t we know.
05:18
So am i missing something or is this problem impossible? it says the tennis ball is dropped, which means that the initial speed is zero because it's dropped from zero.
05:32
So y equals y initial plus v initial t, which should be zero.
05:40
Minus one half vt squared.
05:42
Well, let's just see if this works out, but we're given too much information.
05:48
Y of the collision is 10.
05:54
And we're told, yeah, that the height was 54.
06:04
So we're told 54, no, no, no.
06:12
I'm just substituting into that equation up here.
06:16
10 equals 54 minus 1⁄2 .9 .81 times t squared.
06:35
That's what we're told.
06:38
So let me put this into desmos.
06:42
I'm pretty sure i set this up so that you can see what i'm doing in desmos.
06:49
Not 100 % sure.
06:51
I don't think there's any way for me to tell.
06:54
So i'm just going to put in the right side.
06:57
Of this.
06:58
54 minus 9 .81 over 2.
07:07
No way.
07:20
T -subs c is not 10.
07:24
It's 3.
07:26
At least that'll be a little bit better, but i still don't think it's going to work.
07:46
3 squared is 9.
07:50
Okay, well, it does kind of work.
07:56
All right because we got basically 10 according to the significant digits we're given it doesn't really work but i'd say close enough so now um what are we trying to figure out time of time of its collision with the ball we're already given time what's going on here let me read the question yeah we're actually already given the time of collision but let's go ahead and improve and improve this question, what if we don't know the time of collision? then, rearranging that equation that i put a square around, y sub z minus yt0 equals one half g tc squared.
09:18
Okay.
09:20
Now, i forgot my negative.
09:25
Okay, so now i'm going to multiply by 2 over g, negative 2 over g.
09:32
So i'm going to get 2 y sub t 0 minus y sub c over g equals t sub c squared.
09:46
But now if i just take the square root of it, i can do that.
09:50
Okay, so now if i use that equation, i can see the time when they would collide.
09:58
And it should be really close to three.
10:03
Y sub t0.
10:09
Yt0 is 54, minus y of collision is 10 over g, which is 9 .81.
10:35
Okay, and it is almost exactly three seconds.
10:39
So, all right, maybe i do buy this.
10:42
Because if you round that, you get 3 .00.
10:45
Seconds.
10:47
But again, too much information, we didn't need to be told that at the beginning.
10:52
Okay.
10:54
Philosophy of the rock initially.
10:58
So now i'm going to use the other two equations.
11:06
We know t of collision.
11:10
We know theta.
11:13
And we know x.
11:14
So i don't even need the third equation.
11:16
This is just way too much information.
11:18
If i rearrange the first one, vr0 is going to be x subc over cosine theta times t subc.
11:52
Do i know theta? i guess i really don't know theta.
12:03
So maybe this is not too much information.
12:07
Yeah, i don't know theta.
12:23
Determine the velocity of their rock initially.
12:27
Okay.
12:32
All right, so i'll figure out theta first.
12:35
I'm going to solve both of them for v subr0.
12:40
So y subc plus one half, just solving the next equation, y subc plus one half g tc squared, plus one half g t sub c squared.
13:02
Okay? good plus one half g t sub c squared is equal to vr 0 sine theta t sub c okay vr 0 sine theta t sub c is going to be y c plus one half g t sub c squared okay got it now um vr 0 is going to be r 0 is going to be that over that which from this equation is x sub c over t c cosine theta so now if i multiply by the sine theta oh and i notice these t -subs -c cancel out if i multiply by the sine theta on both sides and i divide by x sub c, i get y subc plus one half g t sub c squared over x sub c equals sine theta over cosine theta, which of course is tangent theta.
15:18
So if i take the inverse tangent of that, then that should give me the answer.
15:26
So y sub c is 10.
15:29
I know that.
15:30
So let's go to desmos.
15:37
By the way, i'm going to call this t...