Calculating Max Height with Energy Conservation

by Patrick Ford
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Alright folks, so in this problem we're giving a ball that's launched at some speed and angle and instead of using the normal projectile motion equations that we've seen to solve for the maximum height, we're actually gonna use energy conservation here. So let's get started. We're gonna have to draw a diagram and set up the problem. Let's do that. So I've got the ground level like this, I've got my ball that's launched at some speed, which is 20 I'm told that the angle that that is launched at. So let's make this a little bit bigger is 37 degrees. So this ball is going to take a parabolic path like this and then it's going to come back down to the ground again. So remember that in these projectile motion problems, we have three points of interest, we have the point A the launch point, the B is the point where it reaches its maximum height. Let's be over here and that's C. Is the point where it reaches, or it goes back down to the initial height. So we want to do is we want to calculate the height here at point B. That's the maximum height. But now we want to use energy conservation to do that. So really what we have is we're gonna have to write an energy conservation equation, that's the second step. Now, if we want to figure out the height at point B, we're gonna have to pick an interval that includes that be so the easiest one to use is gonna be the point from A to B. So we're gonna have to, or the interval from A to B. So we're gonna have to set up an energy conservation equation along this interval here. So this is gonna be K. A. Plus you A plus. Work done by non conservative forces is equal to K. B. Plus you be. So now we're just gonna go ahead and limited and expand the terms. So do we have any kinetic energy at point a. We do because we're told that the initial velocity here is 20. So there's definitely some kinetic energy. What about any potential energy? Remember that's either gravitational potential or spring energy. There's no springs obviously in this problem. And we can do is we can set this point here to be zero. It may not actually be zero. Where we're just gonna call it zero because that's sort of the lowest point in our problem. And that allows us to say that the potential energy here at point A is zero. It doesn't matter. Right? So what about this work done by non conservative? Remember let's say the work that's done by you or friction. There's no you or friction. There's none of those, none of those forces that are acting. It's only gravity. So what about kinetic energy at point B. So, remember here, at point B that the velocity of the object is still just going to is perfectly horizontal at the apex. The we know that the y velocity at the peak is going to be zero but the x velocity is not, the x velocity is gonna be whatever the velocity was at point A it's gonna be V. A. X. This is just equal to V. X. Remember the X component of velocity stays the same throughout the entire projectile motion because there's no acceleration in the X axis. So what happens is there is definitely some kinetic energy here and there's also some potential energy because we're at some height above are zero points. So that's the fourth step. Now we're just gonna go ahead and expand all the terms again and solve. So this is gonna be one half M V A squared plus actually this is gonna be equal to one half M V B squared. Um Plus and this is going to be M. G. And then Y B. So this is actually we're looking for here. One thing we can do is we can cancel the M. S. Because they appear in all the terms of our problems. So really let's go ahead and look through our variables. I know what V. A. Is. I was given that in the problem and the only other variable I have is Y B. That's the target variable. Now I have a. G. Is that's just the constant. The only other variable I need is I need to figure out what's this be, what's this VB which is really just the X velocity at the peak. So in order to do that, I can actually just go ahead and say, well remember that the X velocity is going to be the same throughout the entire problem. So if I'm told what the launch speed and angle is, I can actually figure out what this X component is. So VX is just going to be 20 times the co sign of 37 and that's going to be 16. So that means that this velocity here at the peak is just 16. Not sorry, not squared is just gonna be 16 but it's just going purely to the rights like this, it's lost all of its y velocity but it's still moving with 16 to the right, Alright, so now we can go ahead and plug in. So we have one half, this is gonna be 20 squared equals one half. This is gonna be 16 squared and then plus and this is just gonna be 9.8 times Y B. Alright, so again, we just have to solve for that, we can go ahead and rearrange basically what you're gonna get here. Um Well actually 11 last thing we can do is we can actually multiply this whole entire equation by two. This just kind of gets rid of the one half that makes our equations a little bit easier. So this is just gonna be 20 squared equals 16 squared plus two times 9.8 times yb. All I've done is I've just multiplied times two throughout the entire equation. It just gets rid of the one half. All right, So now all we do is just bring the 16 square to the other side and then we just have to divide by this two times 9.8. What you're gonna get here is 20 squared minus 16 squared, divided by two times 9.8. And this will equal your maximum height. If you go ahead and work the sandwich you're gonna get is a height of 7.35 meters. And that's your final answer. Alright? That's maximum high using energy conservation. Let me know if you have any questions.