Vectors and Projectiles Legacy Problem #31 Guided Solution

This problem falls into the category of what I would call an angled launch projectile. You know that because they give you the angle of launch and so that makes it an angled launch projectile. Originally the projectile, what happens to be a motorcycle jumpster, has initial y velocity and initial x velocity. And I would add that not only is it an angled launch projectile problem, it's a very difficult angled launch projectile problem. In a problem like this, to solve it, you really need to have a good grasp of the concepts. You need to read between the lines as well. What you're given is the dx, the horizontal distance of this projectile, and the angle of launch. Now if you look back to the overview page, you'll notice that there's no equation, no kinematic equations, with the angle of launch. We've got the box y equations, but all of our kinematic equations have Vox and Voy values in them. actually relate the Vox values and the Voy values to the Vo by the theta, the 45 degrees. One thing you do know about this projectile is you know dy equals zero, and you know dx equals 98.3. To say that dy equals zero simply means that it lands where it started. We're told that in the statement of the problem, the landing location, the same height as the launch height. So initially, you could probably start the problem off by saying something like dx equals Vox times t, where the Vox is simply Vocosine theta. So that would lead you to an equation that says 98.3 equals Vocosine of 45 degrees times t. And within that equation are two unknowns, V0 and t. Then you think, oh, maybe I could use a dy equation, and certainly you could. If you pick the equation that goes dy equals Voy times t plus 1.5at squared, then substitute values in. You'd have an equation that looks like this when you're done. It goes zero equal V original times the sine of 45 plus 1.5 times negative 4.9.8 times t squared. Rearranged, it could be written as zero equal V original cosine, sorry, V original sine of 45t minus 4.9t squared. And once more, there are two unknowns, V original and t. So that leaves you with two equations for two unknowns, and any number of ways of manipulating that equation to solve for V original would lead you to the right answer. And you can go ahead and do it that way. That's all I'm going to discuss about that method. I'm going to try something different here. I'm going to talk you through, if you get a sheet of paper and a pen, I'm going to talk you through a derivation of a new equation that can serve quite useful in a situation like this. So it begins with the equation dx equals Vox times t. You can go ahead and write that down, dx equals Vox times t. And then I'm going to take that Vox term and just simply substitute in for Vox. I'm going to substitute in Vox cosine theta, and I'm going to do everything using algebra. So that equation then becomes dx equals Vox cosine theta times t. No substitution in numbers right now. I'm just going to use the variables in the equation. I'm going to try to derive a new equation relating dx to V0. So now when I think about a projectile, I think, well, the time to go up, the time of flight, the t in this equation, is twice the time to go up. And then I think, therefore, I can take 2t up, t up being the time to get up just the peak, and put 2t up into that first equation. The equation then becomes dx equals V original times cosine theta times 2 times t up. Or I could rearrange that to 2 V original times cosine theta times t up. All right. So now we've got an equation that expresses dx in terms of V0, theta, and t up. It would be nice if I could eliminate, reduce the number of variables I have in this dx equation. So I'm going to make an effort to get rid of t up. So here's what I know about a projectile that starts at one location, goes up to the peak, and comes back down. I know the time to go up is simply the original y velocity divided by the acceleration of gravity divided by 9.8. So what I could say about t up is that it is equal to V original sine theta divided by 9.8. And I'll say that again. You should write it down. t up equal V original sine theta divided by 9.8. And I'm going to take all that, and I'm going to put it into the previous equation, the dx equation. I'm going to put that in for t up. What I would then have is dx equal 2 times V original times cosine theta times V original sine theta divided by 9.8. Now I can rearrange this equation into its final form, and that's the equation that would go dx equal V original squared, I've got two of those, times the cosine theta times the sine theta, and then divided by 4.9. Divided by 4.9 comes from, there was a 2 in the numerator and a 9.8 in the denominator. I'm going to say that again. dx equal V original squared times cosine theta times sine theta divided by 4.9. So what have we done? We've written an equation that expresses the dx value, the horizontal distance here, 98.3, in terms of the V original, or the original velocity, and the angle of launch. And you know the angle of launch, and you know the dx, and you could substitute into this equation and carefully solve for the value of dx, and I'll turn it into the previously stated method.