Vectors and Projectiles Legacy Problem #27 Guided Solution

A good problem solver reads the problem carefully and identifies known information, organizing it in a table or a diagram or in some sort of organized form, identifies the unknown quantity, and then plots out a strategy to get from knowns to unknowns. For horizontally launched projectile problems like this one, the best strategy is to list your given information in a so-called XY table, like the one you see here on this web page. As I read through the problem carefully, I identify two numerical quantities, and I'm going to equate those with some of the variables in my kinematic equation. The first quantity, 208 centimeters per second, is clearly a speed, and it's a horizontal speed of a cart that rolls horizontally off a level lab table. So that's V-O-X. V-O-X equals 208 centimeters per second. Now this cart lands a horizontal distance of 96.3 centimeters from the edge of the table, so that is a D-X. I say D-X equals 96.3 centimeters. Now that's all that is explicitly stated in this problem, but there's three bits of information that is implied in this question. The first one is, since this cart is rolling along a table, originally, when it becomes a projectile, the V-O-Y is zero meters per second. So I can list that in the Y column of my table, V-O-Y equals zero meters per second. The other two implied pieces of information are true of every projectile problem. That's that A-X is equal to zero meters per second, and that A-Y equal negative 9.8 meters per second per second. Those two bits of information are true of any projectile, because a projectile is an object upon which the only force is gravity, and gravity accelerates objects at negative 9.8 meters per second per second in a vertical direction. Now my unknown quantity that I'm asked to find here is the height of the table. In other words, what's the vertical displacement of the projectile? Find D-Y. So the basic strategy now involves looking through my X-Y table and looking for three bits of information in one of the two columns. If I want to find D-Y, I'm going to have to use an equation something like V-O-Y times T, no, D-Y equals V-O-Y times T plus one half A-Y-T squared. I need to know T in order to get D-Y. How do you get the T? We go to the three bits of information in your X column. You'll notice that of all the X kinematic equations that we have, the one most often used is D-X equals V-O-X times T. And I know that D-X is 96.3 centimeters and V-O-X is 208 centimeters per second. So I can use that equation to find T. I need to know it in order to calculate D-Y. So plug into that equation and solve for T. When I do it, I get about .4630 seconds. That's the time to move that horizontal distance, and while it's doing that, it's also falling vertically. One note on this calculation of T that I should make, and that is the fact that the quantities V-O-X and D-X are in centimeters, and that's sometimes uncomfortable for us, but the fact that both quantities are in centimeters means I don't really need to convert. The centimeters cancel, and I end up with seconds as my time. Now that I've got the time, I turn to my Y column. I'm going to try to calculate D-Y. D-Y equals V-O-Y times T, and the V-O-Y is zero, plus one-half times A-Y, which is negative 9.8, times T squared, and the T we just calculated. So what I'm going to do is cancel the first term right side of that equation, so that the equation becomes D-Y equals one-half times negative 9.8 times .4630 squared, and I'm going to solve for D-Y. You'll get a negative value, meaning the projectile fell down, but the height of the table is usually as it is, 1.05 meters, with no plus or minus sign.