https://youtu.be/2AxbDlqvi0k
PHYS 1101: Lecture Nine, Part Six
Okay, so we’re ready for step five, and we know that our focus is on V0y. Okay, here’s our two sets of equations, for x and for y, for generic two dimensional motion.
Let me point out that with an acceleration component that’s 0, 0 times t, this term will be 0, and I’m left with an equation that just tells me what I already know, the final horizontal velocity equals the initial.
So this equation becomes pretty useless. The last equation also becomes useless. If this goes to 0, the last terms goes to 0, and again, it just tells me that the velocity doesn’t change. Black this equation out for projectiles.
Let’s look at the middle one. Ax is 0, this term drops out, all I’m left with for a projectile is that the final exposition is equal to the initial plus, the horizontal velocity times time.
This is true for any projectile because Ax is 0. We only have really one equation for the x part of the motion. Final position is the initial plus V0x times time.
Whether we need that or not for this problem, or need to use this equation, I’m not sure yet, but I just wanted to highlight that for you. You will have problems where you probably will need to solve for some aspect of the horizontal motion. This is all you’ve got is this equation to work with.
Okay, but for this problem we need V0y. Here’s our choices. Our three equations.
I paused the video real quick to go grab a copy of the y components that we know, our values that we know. If we want V0y, what options do we have?
Let me add to this that we also don’t know the final time. Given we don’t know the final time, I don’t have enough x information to solve for the time. My first thought is to consider equation three. Let’s see if that would work for us.
What we want is right here, this variable. What variables do we have? We do know the final vertical component of velocity, and we do know Ay, and we know the two y coordinates. We have everything we need to make equation three work for us. That’s what we want to do.
Okay, so it’s a complicated equation, but not too bad. We’ll make it work for us. We’re going to algebraically try to isolate this variable, so I’m going to start rearranging, doing some algebra here. Before I get too far, though, I’m going to kill off as many 0s as I have. I do know that the final y coordinate is 0, and that’s because of where I defined my origin to be.
So, my equation actually looks like this. Oops. I want that to be -2A sub y times y0.
Okay, doing my algebra I’m going to add 2Ay times y0 to both sides. So, I’m getting close here to isolating this. I just now need to undo this square, which means take the square root of both sides. Square root on the left, square root, what I’m left with, on the y on the right, sorry, and I’ve got V0y is equal to the square root of 2Ayy0 plus the y squared.
Now when we go to plug in numbers for this and we take this square root, you have to remember from mathematics that the square root of the number could actually be plus or minus that number, and now that sign has real important physical meaning, and we have to think about which of those signs we want to choose to have this be physically meaningful for the direction we know that that vector has to point.
Let me write this reading left to right, and then we’re going to plug in numbers and see what we get.
Plugging in numbers, I’m going to have the square root of 2 times Ay-9.8 meters per second squared. My initial y coordinate is 3 meters, plus 3 meters.
Now I have to add to this Vy squared. The final y component of the velocity was -8.597.
Notice how carefully I’m always plugging in the real sign and the value for each of these variables.
It’s important because you are always are working with the scaler components here. That doesn’t change just because you’re doing some algebra. The signs in the end are important to the right value and if you end up with a sign that doesn’t physically make sense then that would be a problem.
This negative number when it’s squared, of course, a negative times a negative is positive, this is going to be a negative term here. Let’s see what I get for each of these two terms.
Okay, when I plug this into my calculator, I ended up with a -58.8 for the first term and a 73.91 for the second. This is a good sign to me because I know I can’t physically, or it doesn’t physically have any meaning to take the square root of a negative number, so this better turn out to be positive, and it will, because this is larger than this.
The units also make me happy. These units are the same, because it doesn’t make sense to add these together, and then when I square root a meter squared over seconds squared I’m going to end up with meters per second, which is what I should to have it physically represent a velocity.
Okay. When I do that mathematics or do that final calculation I end up with plus or minus 3.887 meters per second.
I explicitly write the plus or minus because that’s what has to happen. That’s what the square root, I know mathematically, leaves me with, but now let me go back and think which makes physical sense.
The sign has to mean the direction of this vector. What is the correct direction for the V0y component?
I have to go back to my real life physical sketch of the scene. My initial velocity vector is up and to the right and I know that that has to correspond to a positive horizontal and a positive vertical component.
So for the scaler component V0y, I do need to choose the positive value there.
So I need to go with V0y is equal to plus 3.887 meters per second. All right, I’m at the stage now where I’ve used my fundamental equations that have now with their custom variables for this particular problem have led me to a value for a variable that I need to solve the problem. I now need to go back and remind myself specifically what I needed to do with that value.
Okay, I went back to this step where I listed my knowns and the variables I wanted and I remember that I need her initial velocity vector, I need the magnitude and the angle.
For the magnitude, I need the x and the y components to apply the Pythagorean theorem to determine that.
I just scan back up earlier in the problem to remind myself that V0x is 2.303 meters per second. So I can apply both of these to get the magnitude of that initial velocity as being 3.3, oops, 3.887 meters per second squared plus 2.303 meters per second squared.
And so I get, for our magnitude then, which I’ll always for sure take as positive when you do the square root, that’s the physical meaning of the magnitude, I end up with 4.518 meters per second, and I would actually type in then, if this were a WebAssigned problem, 4.52, three significant digits.
So there’s my magnitude. And does that physically make sense? Well, we had an initial velocity that was up and to the right and my vertical part is a little bit bigger than my horizontal part, so let me draw that angle just a little bit steeper here. There’s my Vx and there’s my Vy, and of course I know the hypotenuse should be bigger than either, and that’s reasonable. It looks like it is.
How about the angle of that initial velocity? When we sketch this initial velocity triangle above, and so came up with this equation, you remember we did that based on this being the angle that we’re going after. So this means then that theta is going to be the result of taking the tan inverse. So that’s a function on your calculator, but first you have into your calculator the ratio of 3.887 divided by 2.303.
When I calculate this ratio and then take the tan inverse of it, I end up with 59.4 degrees.
So I would report as the answer to this problem probably something like “The initial velocity with which she left the diving board had a magnitude of 4.52 and she was . . . that was directed at an angle of 59.4 degrees up from the horizontal.”