00:02
To find the horizontal and vertical components, the ball launched at a velocity of 60 feet per second at an angle of 35 degrees.
00:15
The horizontal component is our x component.
00:20
We find that we're taking the magnitude times the cosine of the angle.
00:25
Plugging this in and rounding to one decimal place, we get a horizontal component of 49 .2 feet per second.
00:33
The vertical component or the y component we find by taking the magnitude times the sign of theta.
00:40
In this case, 60 times the sign in 35 degrees, rounded to one decimal place is 34 .1 feet per second.
00:49
So our horizontal component is our x and our power.
00:53
The second question has three separate forces acting at different angles along the x -axis.
01:03
The first is a it's active in 45 degrees and we have a 40 force working at 150 degrees and then finally we have a third force acting at negative 60 degrees and it's 100.
01:34
So the idea is we want to find the magnitude and direction of the resultant vector so if we want to find the result in vector, which i am going to call r, we're going to need to add the x components together and the y components together.
01:53
So i'm going to use a chevron to set this off.
01:55
But the x components will be 50 times the cosine of 45 degrees.
02:02
And to that, 40 times the cosine of 150 degrees.
02:07
And add to that 100 times the cosine of negative 60 degrees.
02:14
The y components are going to look very much the same, but rather using sign as opposed to cosine...