00:01
Hi there, so for this problem, we are told that a project is launched at an angle of 30 degrees and to the horizontal from an initial height that is also given that is 6 .5 feet and with an initial speed that is also given that is equal to 100 feet per second.
00:24
So then with that said for part a of this problem, the question is to find the velocity and position vectors for the time, equal or greater than zero.
00:41
Okay, so the velocity vector for this, we know that that is just simply.
00:46
The first component is the initial speed times the cosine of the angle theta, and the y component for this is the initial speed times the sign of the angle theta.
01:01
So if we substitute the values in here, we will have the following.
01:05
100 times the cosine of 30, degrees for the first component and 100 in the sign of 30 degrees.
01:17
So i don't know if you can answer with this or you need a numerical value.
01:24
Okay.
01:25
So let me give you a numerical value for this as well.
01:29
So the first component i'm going to give you to you with two decimal places.
01:33
That will be 100 times the cosine of 30 degrees.
01:37
So we will obtain 86 .60 for the first component, for the x component.
01:45
And the second component is 100 times the sign of 30 degrees, which is just 50.
01:52
So that's the solution for the first question for the velocity.
01:58
Now, for the position, okay? so the position following from kinemaritz at any given time.
02:08
Oh, sorry, i have made a mistake.
02:10
The velocity for the y component is this expression, but i forgot about the, well, is this minus the acceleration due to gravity times the time.
02:22
So please add that to third because we need the time.
02:28
We know that the x component of speed is constant, but the y component is subjected to the acceleration due to gravity.
02:36
So the acceleration due to gravity that we are going to use in this case is 32.
02:40
So that will be 3.
02:41
32 times the time, okay? so then we need to include in here minus 32 times the time.
02:47
And that's it.
02:48
That's the solution for the velocity.
02:51
For the position of this, that will be.
02:55
The x component of the speed that we already obtained, that is 86 .60, and this times the time, that's it for the x component.
03:04
And for the y component we will have.
03:07
The initial speed that we know is this value right here, which we obtain is 50, and that times the time, but however, we need to include also the initial height.
03:18
So that will be 6 .5 plus 50 times the time, and this minus 1 divided by 2 times the acceleration due to gravity.
03:29
Again, the acceleration due to gravity is 32.
03:32
So from this, we will obtain minus 16 times the time a square.
03:37
So that's it.
03:38
That's a solution for the position back door.
03:41
Now, for part b of this problem, we are asked about to graph the trajectory.
03:49
So, first of all, remember that this and the three graphs that we have, that we are provided as option for the solution of this part of the problem, and we need to account that it is in the vertical component and the horizontal component, okay? so first of all, remember that when the x equals to zero, then we should have that the y component corresponds to the initial altitude.
04:21
So it should start at this point and do something like go up and then go down, okay? just like a projectile.
04:31
So as you can see, the only one that shows this behavior is the option a from the ones that we are getting...