Vectors and Projectiles Legacy Problem #33 Guided Solution

This is what I would call a very difficult angled launch projectile problem. As you read the problem you recognize that there are two quantities that are given and one of them doesn't even find itself inside of our kinematic equations that we typically use. What we have given here is that dx value is 1.8 meters and the angle of launch theta is 35 degrees. This 35 degrees theta is the same theta that we see in our voxvoy equations that go vox equals vo cosine theta and voy equals vo sine theta. But nowhere in our kinematic equations do we see this theta. Now the things that I also know about this projectile that aren't explicitly stated is that I know that dy equals 0, that he lands exactly where he starts. It's a symmetrical trajectory. And then I always know if it's a projectile that ax is 0 and ay is negative 9.8 meters per second squared. I can infer some other things. For instance, I know the time it takes to get to the peak is one half the total time and that the velocity vertically at the peak is 0. There's other things that I know about this problem that's true of any projectile problem of this nature. Now if I want to solve for the vo, I'm going to have to go about it with two equations for two unknowns. So I'll step you through that real quickly. I'm going to use the equation dx equals vox times t. And I'm going to take the 1.8 meters and use that as dx. And I'm going to take the 35 degrees and I'm going to try to figure that into the equation. So the equation would become this. 1.8 equal vo times cosine 35 times t. I can't take it any further than that. There's two unknowns there, vo and t. So I'll say that again. 1.8 equal vo times cosine 35, that's vox, times t. Now I can look for some dy equations. And I do know dy is 0. So I could find the equation that goes something like this. dy equal voy times t plus 1.5 ay t squared. And I can plug some stuff in there and I'd end up with this. 0 equal vo sine 35 times t plus 1.5 times negative 9.8 t squared. That would rearrange to this form. 0 equal vo sine 35 times t minus 4.9 t squared. Where the minus 4.9 is 1.5 times negative 9.8. Now there's still two equations in there. There's vo and t. And those are the same two unknowns from my dx equation. So I have two equations, two unknowns. And theoretically I should be able to solve this equation for vo. I'm going to take a different angle on this problem. You can go ahead and finish the problem that way. You can close this help file and solve for vo. Solve for t first. And then here's the other way to go about it. I'm going to do all of this with variables. No numerical information. I'm going to try to generate an equation that relates dx, vo, and theta. Because here I have dx, I have theta, and I have vo as my unknown. So I'm going to try to generate that equation. It's going to be a universal equation, true for any projectile problem, which a projectile finishes at the same height where they started. So if I do that, I would start with an equation that would go something like this. dx equals vo x t. That's the only goodie. And then I would figure theta in there by saying dx equals vo cosine theta times t. No substitution of numbers now. It's just going to be symbols. And you ought to write that down. dx equals vo cosine theta times t. Now the t is going to be equal to twice the time it takes to get up to the peak. So I'm going to put 2 times t up in for t, and the equation then becomes dx equal vo cosine theta times 2t up. Or rearranged, it would be 2 times vo times cosine theta times t up. Now my goal is to get an equation that has in it dx, vo, and theta, but not t up. So I'm going to try to get rid of t up, and here's how I do it. I focus on the upward motion now of the projectile going from the launch location to the very peak. And during that upward motion, it accelerates at negative 9.8 meters per second squared. And when it finally gets to the peak, the velocity vertically, vo, vfy, is zero. So if I find the equation vfy equals vo y plus ay t, and substitute in zero for vfy and negative 9.8 for ay, and vo sine theta for vo y, that equation becomes this. It becomes zero equals vo sine theta minus 9.8 times up. And I can rearrange it to be t up equals vo sine theta divided by 9.8. So t up equals vo sine theta divided by 9.8. I'm going to stick that expression for t up into my original dx equation, so that that equation now becomes dx equals 2 times vo times cosine theta times, and here's the t up part, times vo times sine theta divided by 9.8. And I've got a 2 in the numerator and a 9.8 in the denominator. I like to simplify that, and I would like to group my vo terms together. So I'm going to rewrite that equation as dx equals vo squared times cosine theta times sine theta, all over 4.9, and there you have it. That's the equation we sought out to generate an equation that relates dx, vo, and theta. And you can use that equation to solve for vo by simply substituting in the theta that you have in this equation, 35 degrees, and the dx value of 1.8.