A projectile is launched from ground level at angle θ and speed v₀ into a headwind that causes a constant horizontal acceleration of magnitude a opposite the direction of motion. a. Find an expression in terms of a and g for the launch angle that gives maximum range.

Video transcript

Hey, everyone in this problem, a missile aspired with a takeoff speed of 120 m per second and an angle beta with respect to the ground, it experiences a horizontal acceleration of 50 m per second squared along the X axis in the direction opposite to the motion. And we're asked to determine the value of beta so that the missile will cover a maximum distance. We're told to use a hint and it is a hint relating the derivative of some sign and cosine functions. And we're gonna come back to that hint as we work through this problem. Now, we have four answer choices all in degrees. Option A 4.55 option B 24. option C 11.1 and option D 5.55. So what we're interested in here is the value of beta so that we cover a maximum distance. Ok. So let's write out all the information we have and see how we can kind of relate those two things. So let's start, we have our missile and it is going to be launched at some angle theater with a speed of 120 m per second. What this tells us is that the initial feed the in the structure, it is going to be equal to that 120 m per second multiplied by cosine of data because that's the adjacent side in the vertical or Y component of that initial speed is gonna be 120 m per second multiplied by sign of data. All right. And that's taking up into the right to be our positive directions. So we have these initial velocities and let's look at the rest of the information we have. So we have some information in the X direction and we have some information in the Y direction. I know in the X direction. We've just written that initial speed. And so the initial velocity because it's in our positive direction will be positive or 120 m per second multiplied by cosine beta. We don't know anything about the final speed. And when the missile reaches that maximum distance, we know that the acceleration in the horizontal direction has a magnitude of m per second squared. And we're told that it's in the direction opposite to the motion. OK. So the motion is to the right, we've said that the right is our positive direction. So the acceleration is acting to the left in the negative direction. And so our acceleration is going to be negative 50 m per second squared delta X. This is gonna be that horizontal distance traveled and we want that to be maximal. OK. So we're gonna call delta XD and D is gonna be that distance that we're trying to maximize. And finally, the time t we aren't told any information about. So what we want here is we wanna figure out an equation for the distance D that we can maximize. However, we only have three variables here that we've written down. OK? We have V on X, we know some information about, we know the acceleration and that's kind of it. OK. We're looking for this delta X this equation for the distance D. So that's not enough known values to use our kinematic or U AM equations. OK. So let's switch over to the Y direction and see what we know there. So again, the initial velocity V not Y 120 m per second multiplied by sin of data. We don't know the final velocity. We know that the acceleration will be the acceleration due to gravity negative 9.8 m per second squared. We know that the vertical displacement delta Y is just going to be m. OK. The missiles fired from the ground and then it's gonna land back on the ground at that maximum distance. And so the vertical displacement, the difference in the initial and final vertical positions is going to be zero. And again, the time to we aren't sure. Now, in this case, in the y direction, we have three kind of known values. OK. We have V not Y we know that other than data, but we can write it in terms of data. We know the acceleration, we know the displacement. So let's use this Y direction information to write an equation for the time P, the time T in the Y direction is gonna be the exact same as the time it takes in the X direction. And so we can use that time T equation then in the X direction to figure out an equation for the distance D that we can then maximize. All right. So let's get started there. And what we're gonna do is we're going to find the time in terms of data. So we're gonna leave the final velocity out. We don't have any information about that. That's not what we're looking for. So we're gonna choose the kinematic equation with the other four variables. And that's gonna be delta Y is equal to V, not Y multiplied by T plus one half multiplied by A Y multiplied by T squared. Substituting in everything we know. 0 m is equal to 120 m per second, multiplied by sine of theta multiplied by T what's one half multiplied by negative 9.8 m per second squared multiplied by T squared. And now we have zero on one side. And then we have this kind of quadratic equation in T on the right hand side. So let's go ahead and factor and we have zero on one side and we factor on the other. We're gonna be able to solve for T. So we have zero is equal to T multiplied by 120 m per second. Sign data minus 4. m per second squared T. Now, on the right hand side, we have this factored equation. So we have one factor multiplied by another factor. If either of those factors are equal to zero, the entire right hand side will be equal to zero and the equation will be satisfied. OK. So either of those are gonna be solutions. So from the first we get that T is equal to zero seconds. From the second, we get 120 m per second multiplied by sin of data minus 4. m per second squared T is equal to zero, which tells us that the TT is equal to m per second divided by 4. m per second squared multiplied by sine of beta. All right. So now we have this equation for T we have to kind of figure out which tea we should be using. Well, in this case, the first is that T is equal to zero seconds. That's gonna be right before that missile is launched. OK. So it's on the ground right before it gets launched. And so that's the initial time point. We don't wanna look at that one, we wanna look at this other time point where it's made it back to the ground. All right. So we're gonna be using that second time that we found and now that we have some information about the time and let me go back to all of the information we have, we're gonna use that in the X direction now to develop an equation for the distance D OK. All right. So now let's find the distance deep, right? In terms. So we're gonna be using the exact same equation again. OK. That kinematic equation where we don't have the final velocity included in this case, it's in the X direction. And so we have a delta X is equal to V not X multiplied by T plus one half multiplied by A X multiplied by T squared. Substituting in our values, we have that the distance D is going to be equal to m per second multiplied by cosine theta multiplied by T which we found to be 120 m per second, divided by 4.9 m per second squared, all multiplied by sine of theta plus one half multiplied by negative m per second squared, multiplied by T squared. So 120 m per second divided by 4.9 m per second squared data. And again, all of that squared and, and those times that we used just come from the time we calculated about OK, the time that it's gonna take for that missile to reach the ground again. All right. So let's go ahead ahead and simplify as much as we can. We have that the distance D is going to be equal to 2900 and 0.7755 m multiplied by cosine of beta multiplied by sine of the minus 14, 0.7526 m multiplied by sign squared babe. All right. So now we've written the distance D in terms of the angle theta. OK. Remember we're trying to find the theta that will maximize the distance, right? How do we maximize the value? Well, what we wanna do is take the derivative and set it equal to zero. OK. So in order to maximize we're gonna take the derivative of D with respect to theta and set it equal to zero. And this is where our hint comes into play. OK? We're told in the hint that the derivative with respect to the of a multiplied by cosine data multiplied by sine data minus B multiplied by sine squared data is equal to a multiplied by cosine of two data minus B multiplied by sine of two data. In this equation on the left hand side is exactly what we're here at. We're taking the derivative with respect to data of some constant multiplied by code multiplied by sine minus some other constant minus sine square or sorry multiplied by sine square. OK. So we can apply this formula in order to find the derivative more simply OK without having to work out all of the messy trig functions. So we have that the derivative DD with respect to D theta will be equal to the derivative with respect to data of 2938. m cosign the sign the minus 14, 0.7526 m multiplied by sine squared data. And again, by the hint, this is going to be equal to 2938. m multiplied by cosine of two data minus 14, 0.7562 m multiplied by two data. OK? So those coefficients stay the same and we get these new trick functions and we want to set this equal to zero. Now, if we set this equal to zero, we can go ahead and solve. So what we have is at 2933. m multiplied by cosine of two data is going to be equal to 14, 0.7562 m multiplied by sign up to data. We're gonna call this equation star and we're just gonna move up where we have more room. We are running out of room. So we are going to move this up on the right hand side here, we have some more space to write. So we're coming back to equation star if we divide both sides by cosine of two data, and then we divide both sides by that 14,993 0.7562. What we end up with is tangent of two data. OK? Because we have sine tooth data divided by cosine of two data that gives us 10 tooth data is gonna be equal to 2938. m divided by 14, 0.7562 m. The unit of meters will divide it. If we take the inverse tangent, we're gonna get that two theta is approximately equal to 11.09 degrees. Then finally dividing by two, gives us that the data is equal to about 5.5 or five degrees. And that's what we wanted to find. We wanted to find the angle theta that maximized the distance and we did exactly that. OK. So that was a very involved question. OK. Let's go back, just refresh what we've done. We wrote out all the information we had in the X and Y direction. We needed to use the wide direction information to figure out the time it takes for the missile to reach the ground again. OK. So we did that we found the time it takes for it to reach the ground. Again, in terms of data, we used that time to find the distance traveled in terms of data. Then we took the derivative with respect to data and set it equal to zero. In order to maximize that distance, we stopped her theta and we found that the angle that's gonna maximize the distance is about 5.545 degrees. If we go up and look at our answer choices, we can see that this corresponds with answer choice. D thanks everyone for watching. I hope this video helped see you in the next one.