Vectors and Projectiles Legacy Problem #5 Guided Solution

A good problem solver has, by nature, a collection of habits that they generally practice. The habits involve reading the problem carefully, looking for cues in the problem that reveal information about the motion, identifying known and unknown information, and then plotting a strategy to get from the known to the unknown information. Typically, it begins by constructing a diagram to get that mental picture going on. And here in this problem, I wouldn't begin to think of solving it until I began by drawing a picture. And my picture would involve a map that shows a person walking from the locker room or from the lunch room to the locker room. I would draw a vector off to the left on my sheet of paper. I call that west. It would be a vector, I don't know how long it would be, two inches or something like that. And I would label it 43 meters. Now, I said it was a vector, which means it has a starting point and a finishing point and an arrowhead pointing in the direction it goes. I usually put that arrowhead at the end of the diagram, at the end of the vector. And then at that point, the person has walked north down the hallway towards the locker room. So I draw a second vector, this one longer than the first since it's a longer displacement. And I draw it straight up on my page of paper, which is north. Put an arrowhead at its end, and I take a moment and I pause. And I ask myself, is that exactly what the problem says the person has done? Walked west and then walked north, and it is. I'm asked to find the magnitude and direction of the resulting displacement. That is, how far out of place is the person from the time of starting to the time of finishing? And so I draw that resulting vector from the tail of the first vector I drew there in the lunch room to the arrowhead of the last vector I drew there in the locker room. I draw the vector diagonally across from the start to the finish, and I look at my diagram now, and I recognize that what I have is a right triangle. And I wish to find the direction of this third vector that's going along the hypotenuse of my right triangle. It has an arrowhead up there near the locker room. Now I want to find that magnitude, so I use Pythagorean theorem, the 43 squared plus the 72 squared equal hypotenuse squared. And I solve that one for the hypotenuse, being careful to follow the orders of operations. Now I have to find the direction of that vector. I could find how many degrees north of west that vector is rotated, or how many degrees west of north, or even how many degrees counterclockwise from east. Any of those conventions generally works. Here I'm going to determine it first north of west. So I notice that inside of my triangle, down at the tail of this hypotenuse, bottom right corner, I notice there's an angle between the leg that's 43 meters long in hypotenuse. And I can find that angle value if I use the tangent function. The tangent function tells me the ratio of the side opposite to the side adjacent. So I just say tangent of that angle equal opposite side over adjacent, or 72 over 43. I find out what 72 over 43 is on my calculator, and then I go second tangent of that ratio. And I get an angle value. It comes out to be a little bit more than 59.1 something, and I round it to 59.2 degrees. That's the number of degrees north of west that that hypotenuse vector is pointing. I could also express that as the number of degrees rotated counterclockwise from west. It certainly is more than 90, as I can tell, since 90 degrees is north. And it's less than 180, since 180 degrees is west. In fact, I've just calculated it's 59.2 degrees less than 180. So I subtract the 59.2 degrees from the 180, and I get the number of degrees counterclockwise from east that this vector is directed. That's all there is to it.