In the absence of air resistance, a projectile that lands at the ...

Question

In the absence of air resistance, a projectile that lands at the elevation from which it was launched achieves maximum range when launched at a 45° angle. Suppose a projectile of mass m is launched with speed into a headwind that exerts a constant, horizontal retarding force Fwᵢₙd = -Fwᵢₙd î. a. Find an expression for the angle at which the range is maximum.

Accepted Answer

Hey, everyone in this problem, we have a hockey player who wants to score a goal and there's a puck with a mass of 0.2 kg. That's initially at rest on the ice. Players gonna hit that puck with their stick giving it an initial speed of 20 m per second. But there's a headwind blowing in the opposite direction of this shot with a constant force of five newtons acting on the puck horizontally. And we're asked to determine the angle at which the puck should be launched to achieve maximum range. When we're given a hint to use the following expression, we're told that the derivative with respect to data of a cosine data sine data minus B sine squared data is equal to a cosine two data minus B sine two data. We're given four answer choices. Option A 30.4 degrees, option B 39.2 degrees, option C 20.4 degrees and option D 10.7 degrees. So let's start by just drawing a little diagram. So we have our puck and it is gonna get shot at 20 m per second, add some angle theta and we wanna figure out what that angle should be to achieve maximum range. So that puck is gonna fly upwards and eventually land back on the ice. And we're gonna call that range or that horizontal distance that the puck travels are. So let's look at the variables we have and we have the X direction and we have the Y direction to consider in the X direction. OK. We have an initial velocity, we know that that's gonna be 20 m per second multiplied by cosine of this angle theta. And we wanna break this 20 m per second into the X and Y components. And the X component is gonna be related to the cosine of the angle because we're talking about the adjacent side. So we have 20 m per second multiplied by cosine up data. OK? And likewise for Y, the initial Y velocity is gonna be 20 m per second multiplied by sine of data. OK? Because it's on the opposite side. Now, we don't know the final velocity in the X direction. We don't know the acceleration in the X direction. We don't know delta X, that's that range. We are calling that R. So we're gonna have that, that's equal to R and we don't know the time that this takes. OK. So we have a lot of unknowns in this situation. If we look in the Y direction, the final velocity again is unknown velocity. When the puck is striking the ground again, we know that we have acceleration due to gravity. OK? And we're gonna take up into the right to be positive. So that acceleration acts down, which is in the negative direction. So we get negative 9.8 m per second squared. We know that the change in position, the change in vertical position, delta Y it's gonna be zero m. OK? When the puck reaches its maximum range, it's back on the ice which is in the same vertical position as it started. So that change in vertical position is equal to zero m. And again, the time T we don't know. So what we wanna do is find the maximum range. So we want to have some expression for R that we can maximize in order to do that, we need more information about the X direction. Now recall that the time T it takes in the X direction is gonna be the same as in the Y direction. OK? That's that same time, whether we're looking at the X component or the Y component. So let's start by looking at the time and let's use the Y direction if we have delta Y and we can write that this is equal to V, not Y multiplied by T plus one half A Y multiplied by T squared. And we chose the kinematic equation that did not include the final velocity because we don't have any information about that. And that's not what we're looking for right now. OK. So let's give ourselves some more room here and we are going to substitute in everything we know. So on the left hand side, we have delta Y which is zero, this is equal to 20 m per second multiplied by sign of data multiplied by T plus one half multiplied by negative 9. m per second squared, multiplied by T squared. We can simplify and we can also factor, can we have this T term in both terms, care or this factor of T in both terms. So we're gonna factor that T out. We get T multiplied by 20 m per second sine data minus 4.9 m per second squared multiplied by T. And we factored one T out. So we still have T left over because we had T squared and now we can solve OK, we have that this is equal to zero. We have two terms multiplied together. So if either of those terms are equal to zero, the entire right hand side will be equal to zero, which will satisfy the equation. And so we get that either T is equal to zero or 20 m per second, multiplied by sin of theta minus 4. m per second squared, multiplied by T is equal to zero. Now solving for tea, yeah, we get that T is equal to 20 m per second multiplied by sin of theta divided by 4.9 m per second squared. So we don't know data yet. That's what we're trying to find. So we can't find an actual numerical value for this T right now. But that's OK. We at least have an expression for T that we can use when we look in the X direction. OK. Now, what else are we told in this problem? What can we use? And we're told about a force, we're told about a force acting in the horizontal direction and recall that we can relate forces to acceleration. OK. So let's use that information to try to find that horizontal acceleration. We're gonna do that on the right hand side here and recall from Newton second long that the sum of the forces is equal to the mass multiplied by the acceleration. And in this case, we're looking in the X direction now that puck is flying to the right, we're told that this force acts on the puck in the opposite direction of the shot. And so the force is gonna be acting to the left in the negative direction. And so we get that the force is going to be negative five newtons that's equal to the mass of 0.2 kg multiplied by the acceleration A X. And so the acceleration in the X direction is going to be equal to negative 25. OK? We have newtons which are kilogram meter per second squared. We divide that by a kilogram and we're left with just meters per second. Squared, which is what we want for acceleration. All right. So we've actually calculated one more piece of information that we can use. We have our time, we have our horizontal acceleration. Let me go back up to the information we have in the horizontal direction. Now, we have a better idea of what we've got. So we have this initial velocity, we know some information about it, but we still don't know theta that's what we're trying to figure out. We now know that the acceleration is negative meters per second squared. And we have some expression for T in terms of data. So let's write an equation for that range R using the information we have now and we're gonna use the same equation we used in the Y direction because we wanna leave out that final velocity that we don't have any information about. We get that delta X is equal to V, not X multiplied by T plus one half A X multiplied by T squared. Delta X is just equal to our range R. So we have R is equal to 20 m per second, cosine of data multiplied by the time which we found to be 20 m per second multiplied by sine data divided by 4.9 meters per second squared. OK? Plus one half multiplied by the acceleration which we found to be negative 25 m per second squared multiplied by the time squared. So again, 20 m per second. Science data divided by 4. meters per second squared, all swear. OK. So this looks really messy but we're just gonna work through bit by bit and simplify. So in this first term, when we look at the coefficients, we have 20 multiplied by 20 divided by 4.9. If we simplify that we get 81.63265, we have meters per second multiplied by meters per second. So meters squared per second squared divided by meters per second squared. So all we're left with is a unit of meters. OK? And then we're multiplying by cosine of data and by sign of data. OK. That's that first term. In the second term, we have one half multiplied by negative multiplied by 20 divided by 4.9 squared. And this is gonna give us negative 0.246564. And again, the units that are left are meters and now we're multiplying by sine data squared. All right. So now we have this expression for the range R, we want to maximize that range. OK. How do we maximize? Well, we want to think about taking the derivative and setting it equal to zero. OK. To find our critical points where a maximum will occur. So we, what we want to find is D R divided by D the and this is where that hint comes into play. Let's go up to our hint. And we are told that if we take the derivative of a cost data sine data minus B sine squared data, we can write it as a co two data minus B sine two data. The expression on the left is exactly what we have. OK? We have this expression with some constant A. OK. In this case, A is 81. multiplied by coast data multiplied by sine data minus some value B which is our 208. multiplied by sine squared data. So we can use that hint and write that the derivative is going to be equal to A which is 81. meters multiplied by cosine of two data minus B 208. m multiplied by sine tooth data. OK? And we want to set this equal to zero to find that critical point or the critical points where the maximum may occur. OK? So we have this set to equals sorry set equal to zero. We can move our 208 over to the right hand side. OK? Then we're gonna divide by that 208 and we're gonna divide by cosine two data. OK? So what we're doing is moving all of the trick functions to one side, all of the constants to the other side. And we get that sign of two data divided by cosine of two data is equal to 81. m divided by 0.24 m. Ok? And I've kept quite a few significant digits here in these intermediate calculations. You always want to check with the textbook you're using or the professor that you have to double check what convention they want you to use when keeping these digits in your intermediate calculations. OK. All right. So sine tooth data divided by tooth data. Well, that's tan tooth data. OK. So what we have here is that 10 2 data is equal to the right hand side this constant value and we can take the inverse tangent to find the two data is equal to 0.41 degrees and dividing by two tells us that data is equal to 10.705 degrees. OK? So let's go through that again. That was a long problem. But we worked it out. We found the angle that we want to shoot the puck at in order to achieve the maximum range. OK. So the first thing we needed to do was find an expression for the time that it will take to reach that maximum range. OK. And we use the Y direction to do that. We found the horizontal acceleration using the forces in Newton's second law with both the time and the acceleration, we can look in the horizontal direction and write an expression for the range R in terms of the angle theta. We then took the derivative and set it equal to zero in order to find the critical point or that data value that gives us the maximum. And we found that that is equal to 10.705 degrees. If we compare this to our answer choices, we see that this corresponds with answer choice. D thanks everyone for watching. I hope this video helped see you in the next one.

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