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. b. What is the angle for maximum range if a is 10% of g?

Video transcript

Hey, everyone in this problem, we're told that a golf ball is hit with a speed of 40 m per second making an angle theta above the horizontal, the ball faces an acceleration along the X axis opposing its motion, which is 15% of the acceleration due to gravity. We're asked to calculate the volume of data so that the ball reaches a maximum horizontal distance. And we're given a hint to use um a particular expression here with the derivative of some trig functions. And we're gonna come back to that when it's time. We have four answer choices all in degrees. Option A 87.8 option B 40. option C 88.7 and option D 43. So let's go ahead and draw a little diagram of what we have first. So we have our golf ball, it's gonna be hit at some angle theta at a speed of 40 m per second. And so we have that our initial velocity in the X direction V not X is gonna be that 40 m per second multiplied by cosine of theta. OK? Because we're talking about the adjacent side and similarly in the Y direction, we have 40 m per second multiplied by sign of data. And with the opposite side, then this ball is gonna reach the maximum distance that we're gonna call. As you can imagine, it's going up and then coming back down. All right. And we're gonna take up into the right to be our positive directions. So what we're interested in is this maximum horizontal distance. So let's write out all of our variables and see what we have now in the X direction. And we've already written out the initial velocity V not X which is m per second multiplied by cosine of theta. We don't know the final velocity in the X direction. We know that the acceleration in the extraction is 15% of the acceleration due to gravity and it opposes the motion. So because it opposes the motion, it's gonna be negative and then it's gonna be 15% of the acceleration due to gravity. So we have negative 0.15 multiplied by G. OK? Remember that G is 9.8 m per second squared. And so if we multiply that by 0.15 we have that our acceleration is gonna be negative 1.47 m per second squared for horizontal displacement delta X is gonna be that value D OK? That maximum horizontal distance we're looking for. And we have no idea how long this is gonna take. So that's all the information we have in the X direction. And we only really have two known values V not X and A X. So we can't just go ahead and solve for this maximum distance or solve for the. There's nothing we can do with this information alone. We need to look at the Y direction as well. So in the right direction, what do we have? Well, our initial velocity Hawaii, it's gonna be 40 m per second, multiplied by sign of data. And we've already mentioned that we don't know what the final velocity will be. The acceleration is gonna be the acceleration due to gravity. So we have negative 9.8 m per second squared. The vertical displacement is going to be 0 m. OK? The ball is gonna start on the ground and when it reaches that maximum horizontal distance, it's gonna be on the ground again. OK? So the change in the vertical position is just zero. And again, we don't know how long this takes. Mm. So in the extraction, we're looking for this maximum distance D in terms of data. OK? We wanna maximize that distance and calculate the value of data, but we don't have enough variables known in the obstruction. So we're gonna look in the Y direction first and we're gonna let the Y direction. So for the time and we're gonna find the time in terms of the angle theta and then we can use that time in the X direction. OK? Because the time it takes in the X direction in the Y direction is exactly the same. So let's go ahead, I'm gonna write this in a different color just so it's easier to keep our equation separate. And we're gonna find the time in terms of data. OK? So we're gonna choose the kinematic or U AM equation that has a V knot, A delta Y and T are the three variables that we know. And the one that we're looking for, and that equation is gonna be delta Y is equal to V, not Y multiplied by T plus one half multiplied by A Y multiplied by T square. Substituting in what we know we have 0 m is equal to 40 m per second multiplied by assign beta multiplied by T plus one half multiplied by negative 9.8 m per second squared, multiplied by T word. We can factor out one of our T terms. So we have zero is equal to T multiplied by 40 m per second. Whoops sign minus 4.9 m per second squared. So now we have two possible values for T and we can see that we have a factored quadratic equation here, we have zero on one side, everything on the right is factored. So if either of our factors is equal to zero, the equation will be satisfied from the first one, we just get that T is equal to zero. And from the second one, we get the 40 m per second multiplied by sine data minus 4.9 m per second squared. Multiplied by T is equal to zero. If we isolate for tea, we're gonna move it to the right hand side and then divide by 4.9 m per second squared. We get that T is gonna be equal to 40 m per second, multiplied by sine data divided by 4.9 m per second squared. All right. So this first T value T equals zero. That is when the ball is on the ground. But before it's actually been hit, OK. This second time point is the time that we're actually interested in. OK. So we're gonna go ahead and use that time in the direction and look at the distance traveled all righty. So let's do that. And we're gonna do that in green just to separate it. Now we're gonna find the distance he traveled in that amount of time. We know that that time corresponds with this horizontal distance, this maximum distance where that ball is hitting the ground again. So let's find that distance. And we're gonna use the same kinematic equation. But in the X direction, we have delta X, which is equal to V, not X multiplied by T plus one half A X multiplied by T square. OK. Substituting in our values we have D is equal to 40 m per second, multiplied by cosine of data multiplied by 40 m per second time data divided by 4. m per second squared. OK. We're substituting in that value of T that we found plus one half multiply a negative 1.7 sorry, 1.47 m per second squared multiplied by T squared. OK. So 40 m per second multiplied by sine data divided by 4.9 m per second squared. All squared. All right. Now, this looks really messy. We're gonna simplify and it's not gonna be so bad. So we have 40 m per sec multiplied by 40 m per second, divided by 4.9 m per second squared. OK. Those are the coefficients in that first trip. If we work that all in our calculator, we're gonna have 0.5306 m. And then we have multiplied by cosine of data multiplied by sine of data. Now, in the next term, we're gonna do this entire square term. OK. Multiplying with our acceleration, we work this out on our calculator and we have minus 48 0. m. OK? And all of that multiplied by sine squared data. So now we have an expression for the distance, the maximum distance in terms of theta, OK. And again, this distance, we wanna maximize OK, we want to maximize d what do we do when we want to maximize something? While we take the derivative, we set it equal to zero OK. So we wanna find the derivative of that distance D with respect to data and set it equal to zero. And this is where the hint is gonna come into play. Let's go up and look at that hint. Now the hint tells us that the derivative with respect to data of a multiplied by cosine data multiplied by sine theta minus B multiplied by sine squared data is gonna be equal to a multiplied by co two data minus B multiplied by sine two data. And this expression on the left hand side is exactly what we have. And we're taking the derivative with respect to data, we have some constant multiplied by co multiplied by sine minus some constant multiplied by sine squared. So we can use that hint exactly as it's written to simplify what we have here. So we wanna find the derivative with respect to the of 326. m multiplied by cosine of data multiplied by sine of data minus 48.9796 m multiplied by sine squared data. And again, according to the hint, this is going to be equal to Yeah. And let me, we wanna set this equal to zero. OK. So this derivative is gonna be equal to 0.5306 m multiplied by cosine of two theta minus 0.9796 m multiplied by sign of the two data. OK. The coefficients stay the same. So this is equal to zero. Now we're getting somewhere and we have assigned data, a coast data. We have this equation just in terms of data and we can finally try to solve for data. So if we move one term to the other side, we have 326.5306 m multiplied by cosine of two, theta is equal to 48.9796 m multiplied by sign of tooth data. OK? And we're gonna call this equation star and we're just gonna move up where we have more space to continue writing. Now, from this point, if we divide both sides by cosine of two beta and then we divide both sides by this 48.9796. We're gonna have sine two data divided by co two data. OK? Well, sine divided by codes of the same angle. That's the tangent. OK? So we're gonna have 10, 2 data on one side. And then we're gonna have this division of our coefficients on the other side. So let's go up where we have some more space and do exactly that. OK? And I'm just gonna put this start that we're continuing from, OK? And we get hand too. Data is equal to 0.5306 m divided by 0.9796 m. OK? We can take the inverse tangent of this and we're gonna get that two, theta is equal to about 81. degrees, dividing by two, gives us that theta is equal to 0.735 degrees. OK? And that is the angle where we're gonna have the maximum distance. That's exactly what we were looking. OK. So this was a, a very involved problem. OK. Let's walk through what we did. We wrote down all of our information in the direction and the wide direction we found the time it would take for the ball to go all the way up and back down. In terms of the angle data that we were interested. Then we wrote out the distance. OK? We wrote the distance traveled in that amount of time. We took the derivative and set it equal to zero to maximize the distance. And then we solved for the, if we compare the theta value, we found to the answer choices, we're going round to one decimal place and we can see that the correct answer is option B 40.7 degrees. Thanks everyone for watching. I hope this video helped see you in the next one.