Vectors and Projectiles Legacy Problem #22 Guided Solution

This is a class of projectile problems known as a horizontally-launched projectile problem in which the projectile is actually starting its motion with no vertical velocity, completely horizontal velocity. We know this because our little swimmer is running off a high-dive board. In other words, the board goes horizontally, he's running along it, so initially there's no vertical velocity. Now, I recommend that you use the provided x-y table in order to organize your given information before you begin plotting a strategy to get from the given to the unknown information. So here, what I can reason in the x-column of our x-y table is that the Vox is 4.6 meters per second. That's the original x-velocity. I can always reason for projectile problems that the Ax is 0 meters per second. In the x-column, that's all I know originally. I'm eventually going to try to find the dx value. That's the part b question, so that's one of my unknowns. In the y-column of the x-y table, I know several things, one of which is explicitly stated. I know that the projectile, our little swimmer, falls 2.3 meters vertically from the board to the water. I can say that dy equals negative 2.3 meters. There are also two things that I know that are implied here. The first one is that Voy is 0 meters per second. In other words, it's a horizontally launched projectile. So I write Voy equals 0 meters per second. And the thing I always know about every projectile is that Ay equals negative 9.8 meters per second. Now, if I look at what I have listed in my x-y table, I have three bits of vertical information and two bits of horizontal information. Whenever I have three bits of information in one of the two columns, I can calculate a fourth bit of information. So if I focus on my y-column, I can use that information to calculate t as long as I find the appropriate equation. If you need to go back to the x and y kinematic equations that are listed in the r-view page to this set of problems and find one that is useful for calculating the t, that's the first question, a, determine the time to fall 2.3 meters. Now, one of the equations there is Dy equals Voyt plus 1.5 Ayt squared. Now, if you look at that equation, you know that Dy is negative 2.3, you know that Voy is 0, and you know that Ay is negative 9.8. The fact that Voy is 0 means that you can cancel the first term on the right side, and the equation reduces to negative 2.3 equals 1.5 times negative 9.8 times t squared. In other words, negative 2.3 equals negative 4.9 t squared, and you can solve that one for t, and if you do, you should get about 0.6851 seconds. Now that you've found the t, you can use that t as a scalar in either side of the equation, so I'm going to use that to calculate the answer to part b, which is find dx. If you think about the equation for calculating dx, there's only one use, and that is that dx equals Vox times t. The equation tells us that we need to know two things to calculate dx. We need to know Vox, which is given as 4.6, and we need to know t, which we just calculated to be 0.6851 seconds. So plug those two values into your dx equation, and now you can solve for dx. When I do that, I get about 3.1515 meters as my value for dx. Take a deep breath, we've got some thinking to do now for part c. In part c, we wish to find the speed at which the projectile hits the water. Now if I think about the question, find the speed, they're asking me to find the magnitude of the velocity at the very end of the projectile motion, to find that magnitude of the velocity at 0.6851 seconds after launch. Now I need to draw up some conceptual understanding of projectiles, the first idea being that a projectile never changes its horizontal velocity. It only changes its vertical velocity. So at the point that Matthew hits the water, he's moving 4.6 meters per second horizontally, and he's moving with some vertical velocity, which I'll have to calculate. I need to take these two components of his velocity and add them together to get the final velocity, the magnitude of which is the answer to part c. So I'm going to draw a little velocity diagram. I'm going to draw one vector horizontally. I'm going to label it 4.6. I'm going to add two at a second velocity vector vertically. I'm going to call it Vfy. I need to calculate Vfy using one of my equations from the overview page. The one that is of great use here is that Vfy equals Voy of 0 plus Ay times t. So if I substitute negative 9.8 into that equation for Ay, and 0.6851 seconds from part a answer, I can calculate the Vfy. When I do that, I get a Vfy of 6.7142 meters per second. And now I can find that Vf, the velocity at a diagonal direction, is simply the hypotenuse of a right triangle that has its sides, 4.6 meters per second and 6.7142 meters per second. If I apply Pythagorean Theorem to that, I get 8.3188 meters per second. I can round that and that would be my answer.