Vectors and Projectiles Legacy Problem #26 Guided Solution

A good problem solver has a habit of reading the problem carefully, identifying known and unknown information, listing it, and then finding a strategy to get from the known information to the unknown information. For problems like this, horizontal launch projectile problems, it is recommended that you organize your information in an XY table, such as the one shown on this web page. So if I read through the problem carefully, what I observe is that there are two quantities which are explicitly given. What I need to do is equate those quantities with some variables that are typically used in our projectile equations. The first quantity I see is the quantity 1.25 meters per second. I recognize that to be a speed by its units, plus it says it's a speed. And I recognize it to be a horizontal speed, because my picture of the situation is there's a train moving along a countertop and a french fry on the top of the train. And that little french fry falls off of the train, moving horizontally, and then free falls to the ground below. The 1.25 meters per second is the VOX, and I'll list it as such in my X column in my table. The other information that I have explicitly stated is this 113 centimeters. The 113 centimeters represents the vertical falling distance from the top of the train to the floor below. The first thing you have to notice about that 113 centimeters is that's a pretty nasty unit. It doesn't match real well with some of the other units that we use, so I'm going to convert that to 1.13 meters, and I'm going to list that as the dy value. So in the Y column of my table, I list dy equal 1.13 meters. Those are the only two quantities that are explicitly stated in this problem, but there are three quantities that are implied. The first one we'll talk about is the VOY value. It's zero meters per second, since originally that french fry is not moving vertically. It is only moving horizontally, so originally when it becomes a projectile, the VOY value is zero meters per second, so I list it in my table. The other two quantities, which are always true of any projectile, is that AX equals zero meters per second per second, and AY equals negative 9.8 meters per second per second. So now I have five bits of information listed in my table. They're asking me to calculate three things. The first one is the time to fall, the second one is the horizontal displacement, or DX, and the third one is the speed at which it strikes the floor. Now the strategy typically involves looking into your two columns to look for three bits of information, either in a horizontal column or a vertical column, and here we have three bits of given information in our vertical column. The value of having three is that all of our kinematic equations have four quantities in them. So if you know three, you can calculate a four. The first question is what's the time to fall? If I look back at my list of kinematic equations, one of them reads DY equals VOYT plus one-half AYT squared. The DY is 1.13 centimeters negative, the VOY is zero, and the AY is negative 9.8. The fact that VOY is zero means that that first term on the right side of the equation cancels out and the equation simplifies to negative 1.13 meters equal one-half times negative 9.8 times T squared. I can solve for T here, and when I do solve for T, I'm going to get a value of 0.1519 seconds. Now that's my time to fall vertically from the top of the train to the floor below, and while it's falling vertically, it's also moving horizontally. We're told here at a speed of 1.25 meters per second. So in part B, to find the horizontal displacement, you use the equation DX equals VOX times T, where the VOX is 1.25 meters per second and the T is 0.1519 seconds. That gets you the DX value. Take a deep breath because this part C is difficult. In part C, they're asking you to find the speed at which the french fry strikes the floor. Find the speed, the final speed of the french fry. It's not asking you to find VFX nor VFY. It's asking you to find VF. It is find the vector sum of the VFX and the VFY. So I'm going to set myself up a little triangle with the VFX going horizontally and then adding to that the VFY going vertically and drawing the resultant, which is the hypotenuse of a right triangle that has as its side VFX and VFY. Now the VFX is easy to determine. It's the same as the VOX I draw from my understanding of projectiles. To get that, so VFX is 1.25 meters per second, VFY I'll have to calculate. I'll have to calculate VFY from the idea that it's accelerating vertically at negative 9.8 meters per second for 0.1519 seconds and I'll have to go VFY equal the VOY of zero plus AY times T. VOY cancels and so VFY is negative 9.8 times the 0.1519 seconds. That's going to give me a value of 1.4882 meters per second for VFY and I can use that in my calculation of the final speed. Applying Pythagorean theorem, I would go 1.25 squared plus 1.4882 squared is equal to VF squared and I would solve for VF.