The cannon in FIGURE CP4.83 fires a projectile at launch angle θ with respect to the slope, which is at angle Φ. Find the launch angle that maximizes d.

Hint: Choosing the proper coordinate system is essential. There are two options.

Question

The cannon in FIGURE CP4.83 fires a projectile at launch angle θ with respect to the slope, which is at angle Φ. Find the launch angle that maximizes d.

Hint: Choosing the proper coordinate system is essential. There are two options.

Accepted Answer

Hey, everyone in this problem, a gardener is watering his plants on land inclined at 20 degrees, which is shown in the figure, the water is leaving the pipe at an angle of 30 degrees with respect to the slope. And we're asked to determine the angle at which the water should leave the pipe. So it covers a maximum distance deep. And we're given this hint that has some trick functions and it has a derivative. And we're gonna come back to that, that as we work through this problem, we have four answer choices all in degrees. Option A 70 option B 35 option C 45 and option D 25. If you look at our diagram here and we have this slope making it an angle of degrees with the horizontal, we have our pipe with the water leaving at an angle of theta degrees and then covering this maximum distance D OK. And we want to find this angle theta in order to maximize that distance. Let's think about all of the information we have. We're gonna write out our variables for both the X direction and the Y direction and see what we're able to do here. So in the X direction, our initial velocity OK is gonna be some value V. That's gonna be that initial speed if we look at the particular angle multiplied by cosine of that angle theta. OK. Because the X, the horizontal speed is going to be that adjacent side. And similarly, for the Y direction we have the V nought Y is going to be the speed V. Not that we don't really know yet multiplied by sign of data. All right. So we don't know the exact numerical value of those initial velocities, but we have an idea. And while we're writing out this information, OK, what you'll notice is that I've written the velocities as they relate to this slope. OK. So we're taking the X positive X direction to be parallel to that slope moving to the right and the Y direction, the vertical direction is gonna be perpendicular to that. And so we've kind of tilted our axis. Now, the final velocity in both cases, in both directions, we aren't given information about. We don't know what the speed will be when it reaches a maximum distance for our acceleration. And we have a horizontal and a vertical acceleration and that's gonna come from the force of gravity. And we have to be a little bit careful here. Typically the acceleration due to gravity, we write that in the y direction negative 9.8 m per second squared. And we move on. But in this case, we've tilted our axis. OK. So the acceleration due to gravity, the force of gravity is going to act straight downwards which is actually at an angle of degrees because of our um tilted land. So the acceleration in the horizontal direction and we can see that that's actually gonna be the opposite side in this case. So that's gonna give us a negative side of degrees multiplied by the acceleration due to grab eg and in the Y direction, it's gonna be negative cosine of 20 degrees multiplied by that acceleration due to gravity. G. Now the horizontal displacement delta X is just going to be that distance D that we're trying to maximize in the Y direction, the displacement is going to be 0 m. Mhm If we consider our tilted axis, the water starts and ends at the exact same vertical position. OK. So that change in vertical position is just zero. And the last variable we have in both cases is the time T, OK? And remember that the time T is gonna be the same in both directions. OK. So we don't know that value yet, but we do know that it's exactly the same for both directions. So what can we do here? We wanna find data so that we maximize the distance in order to maximize the distance D we need to have some equation or relationship that represents the distance. OK. Now we only have two other values in the X direction. So that's not enough to use our U AM or kinematic equations. So let's start with the Y direction. OK. We have three kind of known values or three values that we at least have some information about. We can use that to calculate the time T in terms of those other variables. Then we can use that time T to write this distance do so that we can maximize it. OK. So we are going to start by finding she in terms of those unknown variables V and beta. OK. So again, finding T in terms of von data working with this Y direction information. So we're gonna choose the U AM or kinematic equation that does not include the final velocity because we don't have information about that. We're not interested in that. And the equation is going to be delta Y is equal to V not Y multiplied by T plus one half multiplied by A T squared. Substituting in information 0 m is equal to V sign of data multiplied by T plus one half multiplied by negative cosine of 20 degrees multiplied by 9.8 m per second squared multiplied by T squared. So now we have a zero on one side and this quadratic function for T on the other side, hey, don't get confused with this sign data. The von, remember we're looking for T we have a T term and A T squared. Term added together the quadratic eli. So let's go ahead and factor on the right hand side and we can take out a factor of tea from both terms. So we factor up that T and we're left with Vena multiplied by a sign of data minus 4.9 m per second squared multiplied by cosine of 20 degrees multiplied by T. Now we have this factored equation equal to zero. If, if either one of those factors is equal to zero, the entire right hand side will be equal to zero. And our equation will be satisfied. OK. So if we set our first factor equal to zero, we get that T is equal to zero. If we said that second factory equal to zero, we get V not signed data minus 4. m per second squared, cosine of 20 degrees T is equal to zero. We can subtract our V not sine data and then divide by negative 4.9 m per second squared cosine of 20 degrees to get that the time T is going to be equal to V. No sign of data invited by 4.9 m per second squared cosine of 20 degrees. And we can simplify that denominator and approximate it. So we have V no sign of the divided by 4.6045 meters per second squared. All right. So step one is done. We figured on the time it's gonna take the water to reach this maximum distance in terms of these other variables we have. Now let's look back at our information in the X direction now that we have some information about the time T, we can use that to write an expression for the distance T, OK. So we're gonna do that, we're gonna write D in terms of those other variants. All right, now we're gonna use the exact same kinematic equation as we did in the Y direction because again, we're not interested in the final velocity. It's those other four variables that we want. So in the X direction, that's gonna be delta X, it's equal to V, not X multiplied by T plus one half A X multiplied by T squared. Substituting inner values we get that D is equal to V knot cosign of data multiplied by the time we just found, OK. And we found two times, we have to figure out which one to use. Well, the first one is just time zero. OK, zero seconds. And that's right before the water is actually coming out of that hose or the pipe. OK. So that's not the time we want to use. We want to use that second time when it actually reaches that maximum distance. OK. So we're gonna use that second time Veno sign data divided by 4.60 or 5 m per second squared. Then we add one half multiplied by the acceleration, negative sign of 20 degrees multiplied by 9.8 m per second squared and multiplied by T square. So V not sign up data divided by 4.6045 m per second squared and all of that is squared. All right. So this looks really messy, but we're gonna simplify, we're gonna clean it up. We're gonna kind of combine all of these coefficients together. So we're gonna have that the distance D is equal to V not squared. OK? Remember that V naught is a number. We don't know what the number is, but it is just a number. OK? So we have V not squared divided by 4.60 meters per second squared multiplied by cosine of data. Sign of data minus 0. 65. And the units here will be second squared per meter multiplied by V not squared multiplied by sine squared of data. All right. So we found an expression for the distance D in terms of data. OK? So it's some number multiplied by coast multiplied by sine minus some other number multiplied by sine squared. And what we wanna do is we wanna maximize the distance D. OK. So the last step here is we wanna maximize D how do we do that? How can we maximize an expression or an equation? OK. Well, what we're gonna do is take the derivative with respect to the variable we're interested in data and set it equals to zero. And that's how we're gonna maximize it. And this is where that hint comes into play. Let's go back to that hint. And the hint tells us that if we take the derivative with respect to data of a multiplied by coast data multiplied by sine data minus B multiplied by sine squared data, we can write it as a multiplied by co two data minus B multiplied by sine two data. So the left hand side is exactly what we have. OK. With A and B being replaced with these coefficients, these numbers that we have. OK. So this gives us a quick, simple way to write this derivative without having to do all of the in between steps. That's simplify. So let's go ahead and use that. We are looking for the derivative with respect to theta of D. That's gonna be the derivative with respect to data of this entire expression. We found earlier V not squared divided by 4.6045 m per second squared. Post data science data mine is zero 0. second squared per meter V not squared nine squared theta. All right. Now remember what I said V is just a number. OK. So that's gonna be included in that A or B coefficient. OK. So when we take the derivative using the hint, we get that this is going to be equal to V not squared divided by 4. 4 5 m per second squared, multiplied by cosine of two data minus 0. 2nd squared per meter. V not squared multiplied by sine of two theta. And we're gonna set all of this equal to zero. All right. So we've simplified and what you'll see now is that we can divide both sides by V not squared. OK? And that V not squared term is going to go away. So when you're starting a question, if you have a value like that, that you just, you don't know, don't freak out. OK. Work through the problem. There's a good chance that that value may divide it at some point and you don't need to worry about it. OK? So work with it just as the, not like we've done here and then we've divided it out and we no longer have to worry about it. So we're really, really close. Now, we have this equation, the only unknown in this equation is data and that's perfect. That's exactly what we wanted to find. We wanna find data that maximizes its distance D. So now it's just a matter of solving for data. So we can move our sine two data term to the other side by adding, then we're gonna divide both sides by cosine of two data, divide by the coefficient on the sine two data. And what we'll get is a sign of two data divided by cosine of two, theta is equal to one divided by 4.6045 m per second squared, multiplied by 0. 2nd squared per meter. And the units are going to divide it in that denominator. Now we have sock tooth data divided by coast tooth data. Recall that that's just tangent. OK. So the left hand side, we have two or sorry tan tooth data. On the right hand side. If we simplify, we get that this is equal to 2. we can take the inverse tangent to find that two theta is about 70 degrees and dividing by two, we get that data is equal to 35 degrees. All right. And that's what we wanted. We wanted a data value that maximized our distance. So that was a very involved question. Let's recap really quickly. OK. We started by writing out all of our information in the X and Y direction. We needed to find the time T in terms of data using our Y direction information so that we could write the distance D in terms of data. And we needed the time in order to do that. Once we wrote an expression for the distance D, we maximized it by taking the derivative and setting it equal to zero. And then we were able to solve for the corresponding value of data. OK. So we found that the is about 35 degrees. Going to our answer choices. We can see that this corresponds with answer choice and b thanks everyone for watching. I hope this video helped see you in the next one.

The cannon in FIGURE CP4.83 fires a projectile at launch angle θ with respect to the slope, which is at angle Φ. Find the launch angle that maximizes d. Hint: Choosing the proper coordinate system is essential. There are two options.

Verified Solution

Key Concepts

Projectile Motion

Coordinate Systems

Optimization in Physics