https://youtu.be/F6kdYcB9qkg
PHYS 1101: Lecture Nine, Part Five
Now let’s move down to step number three. Let’s use our trigonometry and calculate as many components as we can. Well we’ve already stated what the components are for the acceleration vector. The only one we need to do some trigonometry on, and were able to, is this final velocity vector because I do know the magnitude in this angle.
So let’s draw that down here. I have a pretty steep triangle. That’s an excellent triangle. I know that this is a 75 degree angle. I need to get an angle inside of my right triangle to do my trig. This is 90 degrees so I know that this has to be 15. 75, 85, 90; that adds up to 90 degrees. So here is V, the magnitude of my final velocity vector, her speed when she hits the water; I need to bet the magnitude, Vy and the magnitude of Vx from this geometry.
Well, Vy, that’s the adjacent side. So for this scenario Vy is going to be V time the cosine of 15 degrees, meaning I have to calculate her speed when she hit the water which was 8.9 meters per second times the cosine of 15 degrees. And keeping four digits for that I get 8.597 meters per second. That’s Vy.
And is it positive or negative? Why don’t I take the time right now to think that through. The hypotenuse when she hits the water, that velocity is down and to the right so Vy has to capture the downward trend; has to be negative. So I am going to write this as minus 8.597.
It doesn’t matter when you decide to think about the right sine other than when get down to making your official list of knowns, for sure at that point, it’s the scalar components that you’re writing down and you have to have the right sine then so here is the right sine at this point. Vx, that’s the opposite side; I need to use the sine to calculate that. So that’s 8.9 meters per second times the sine of 15 degrees and that gives me 2.303 meters per second. Note the four digits for these intermediate calculations.
What’s the right sign for this? Vx has to capture the rightward trend of this vector. That’s why the vector’s sum does add up to the components equaling the hypotenuse. And this component points in the positive direction because of course to remind you, you’ve chosen the standard axes directions. So that’s really done with step 3. We’ve calculated as many components as we can, in addition to knowing what Ax and Ay are which are givens because she’s a projectile.
So now let’s go down to step 4 and let’s make our complete list of the knowns and the variable that we need to solve for. If we do this right when we go to look at our equations it’s just so much easier to see it summarized so much clearly; what we have and what we need. So for x. x0 was zero. Ugh, I can’t back it up to have that on the same page.
The final x position we didn’t know. The initial x component of the velocity; I am going to pause for a minute here. Then I am going to go, I got position, velocity, start and finish, and then acceleration. Ax, I know is 0. Vx, I just calculated, is 2.303 meters per second. Okay, if Ax is 0, my velocity can’t change. If it was this at the end of the problem then sure enough that’s what it is at the start of the problem, just as she leaves the board.
Let’s go over now to the Y variables. Her initial Y coordinate, we decided to put our origin at the water level, so her initial is plus 3 meters. Her final is then 0. The initial y velocity component we don’t know but we did just calculate the final y component of the velocity and that was minus 8.597 meters per second. The acceleration in the y direction is minus 9.8 meters per second squared.
Okay. Between the two components I am going to sketch in; I know I have to find my initial time to be zero and what does the clock read then when she hits the water? Don’t know. We weren’t given that.
The other thing I need to do for this step is to decide what variable I clearly need to focus on. Let’s go back to the, the problem and read again and see what we need to answer or solve for this problem. They want us to calculate her initial velocity, both the magnitude and the direction. We need the complete vector information for this initial velocity vector.
What does that translate to for the variables that we have, the 12 variables that we’re forced to work with in our equations that will we use. I need the horizontal and the vertical components of the initial velocity. With those I can calculate. Let’s do a quick sketch here. V0, V0x, and V0y. If know the magnitude of these two vectors, the two sides, I can use my Pythagorean to get the hypotenuse. That will be the magnitude of her initial velocity and then I am also going to have to figure out the angle.
So in terms of variables to focus on, I need, first, to solve for the variables Vx, V0x, and V0y. V0x is a known; I have that value so what I really need is this, V0y. Once I have that, those, then I going to get the magnitude by using my Pythagorean theorem. V0x squared plus V0y squared and then I can use my trigonometry to get the angle. I would, might as well, go with this angle and then I can report that direction as being so many degrees up from the horizontal.
So how would I get this angle? Well tangent is a good . . . I can use any trig function. I’m going to use tangent. Tangent of this angle, let’s call it theta. Tangent of theta is equal to opposite over adjacent. So this will just be the magnitude of the vertical component and the horizontal. And to solve for that angle I take the tan inverse of both sides. Tan inverse of the left gives me theta and then I need take the tan inverse of the right. So that’s what I am going to do once I get values for both of these. I’ve got V0x; let’s go after V0y. That’s the name of the game.