00:01
So in this problem, we have a rocket with a mass m equal to 4 .13 times 10 to the 5th kilograms.
00:20
And so let's do a rough sketch here.
00:23
So here we have our rocket, and we are told that there is a thrust force, which i'm going to know as an arrow, and f.
00:31
And that thrust force is applied with respect to a theta equals 69 .7 degrees above the horizontal.
00:44
And i guess while assessing the force is acting our mass, let's go ahead and write down the force of gravity, which we can say is equal to mj.
00:55
Now in the first part, we want to find the magnitude of acceleration.
01:07
The acceleration magnitude there and then the second part we want to find the direction of our acceleration vector so what we're going to do is apply a new in second law so we're going to apply f equals m a but for respective components so we're essentially going to calculate let me just write this with respective components we're essentially going to calculate our acceleration in the x direction and our acceleration in the y direction.
01:51
So how do we do that? well, we can rearrange our formula for f equals ma, and we can say that a sub x is equal to f of x over m, f of x being the horizontal force of the thrust.
02:11
So then f of x, looking at our angle theta, you can say f of x is equal to f times cosine of theta over our mass.
02:25
So computing this, we get that, and actually i should mention our thrust forces given in the problem with a magnitude of 8 .41 times 10 to the 6th newtons.
02:40
And so we're going to have for our acceleration in the x direction, we're going to first compute the horizontal force, which is 8 .4 times 10 to the 6th times cosine of 69 .7.
03:01
And this will all be divided by our mass, which is 4 .13 times 10 to the 5th.
03:10
And we get a corresponding acceleration in the x direction equal to 7 .0667 meters per second squared.
03:24
So that's one component.
03:26
Now let's do the same for our acceleration in the y direction.
03:31
So a y will be equal to f y over m, and our f y will actually be a series of two forces.
03:39
First, we're going to have the positive y force, which is our f, sine of theta, but we're also going to subtract the force of gravity acting on the rocket, because the force of gravity is actually trying to pull the rocket downwards, and the negative y direction, and it will still be all over the mass.
04:03
So now this will be equal to 8 .41 times 10 to the 6 times sine, and of 69 .7.
04:19
I'm just going to go ahead and bring this down.
04:22
So i have more space.
04:24
And so we're going to have the terms...