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Ch 04: Kinematics in Two Dimensions

Chapter 4, Problem 4

A cannonball is fired at 100 m/s from a barrel tilted upward at 25°. c. What is the angle after the cannonball travels 500 m?

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Hey, everyone in this problem, a football is kicked from a cliff with a takeoff speed of 30m/s and a launch angle of 35°. We're asked to calculate the angle Of the ball after it has covered a horizontal distance of 75 m. We're given four answer choices. Option A negative 27.3 degrees. Option B 27.3° option C negative 51.7 degrees And option D 51.7°. Now let's just draw a quick little diagram to get a sense of what's going on. So we have our football and it's being kicked, OK, with some angle At a speed of 30 m/s. And we're told that that angle is 35°. So we have this angle to the horizontal Is 35°. All right. So in this problem, let's take up into the right to be our positive directions. OK? So that initial direction of the ball is going to be positive. Now, we're asked to find the angle of the ball in order to find the angle, we're gonna need to break this into our X direction and our Y direction, we're gonna wanna find the velocities in each of those directions. OK. Recall that velocity contains information about the magnitude and the direction. OK. So that direction piece is the key component that we're gonna be looking for in order to find the angle. OK. So in the X direction, what do we have? Well, let's write down all of our variables. We know that the initial velocity in the X direction Is something that we can find. OK. We're given the initial speed and the angle. So the velocity in the X direction is gonna be related through the cosine because it's on this adjacent side. And so we get cosine of 35° multiplied by 30 m per second. OK. That's our initial speed in the X direction. Now, we aren't given our final speed or velocity in the X direction, but that's what we want to find. We want to find this final velocity in the X direction. That's gonna help us calculate that final angle. The acceleration in the X direction is going to be zero m per second squared. OK. We're gonna assume things are nice. We don't have air resistance. We know that we're gonna have gravitational acceleration, but that acts straight downwards. So that's gonna act only in the Y direction, not in the X direction. Now, we also know that the ball covers a horizontal distance of 75 m. So Delta X, OK. The horizontal displacement is going to be 75 m. And we aren't given any information about the time t now, we want to find this X direction velocity. So we can do that. We have three knownn values vnat X A X and delta X. So we can use those three values to find this velocity in the X direction. OK. So let's go ahead and do that first and then we're gonna come back to the Y direction. So finding the X velocity, we're gonna choose the U AM or kinematic equation. That includes those four variables, the three that we know and the one that we're trying to find, and that's gonna be the following V FX squared is equal to V not X squared Plus two A x multiplied by Delta X. OK. Substituting in what we know the acceleration in the X direction is zero. So the second term goes to zero, OK. We get that V FX squared is equal to Coast of 35° Multiplied by 30 m/s, all squared. OK? We can take the square root on both sides and we get that our final velocity in the X direction Is equal to just cosine of 35° multiplied by 30 m/s. OK? Which is equal to 24.57. Let me write this down below. I don't want to run out of room for the Y direction. This is equal to 24. m per second. Ok. And what you'll notice is this final velocity we found in the X direction is the same as the initial velocity. And we said there was no acceleration acting in the X direction. So there's nothing that's going to speed up or slow down that ball in the X direction. And so it's going to be the exact same as what we started with moving to the Y direction. And remember we need to find the velocity in both directions. In order to calculate that angle in the Y direction. We have the initial velocity a similarly to in the X direction. But in this case related through sign of the angle because it's the opposite sign side, sorry. So we have sine of 35° multiplied by 30 m/s. The final velocity in the Y direction is what we're looking for. In this case, we know that the acceleration is gonna be the acceleration due to gravity which acts downwards, which is our negative direction. So this is gonna be negative 9.8 m per second squared. Now, we aren't given any information about the height of the ball after it's covered this horizontal distance of 75 m. So we don't know delta why and we don't know the time T so in this case, we only have two known values. One thing we're trying to find that's not enough to use our kinematic or U AM equations, we need three pieces of known information in order to calculate V F Y. Now, what can we do here? Well, we know that the time in the Y direction is the same as the time in the X direction. OK. This ball is at the same instant when we're looking at the horizontal and vertical components. So the time is gonna be the same. So if we can calculate the time using the X direction information where we have more known values, we can use that in the Y direction to calculate V F Y. So let's do that, let's find the time. OK? And we're gonna do this with our X direction. Now we're gonna choose the equation that includes time and we can choose any equation because we know all of the other values. Now, so we're gonna choose delta X is equal to V, not T plus one half A T squared. And these should be, this should be A X and V, not X right now, we know our acceleration is zero. So this second term goes to zero, substituting in everything else we know we have 75 m is equal to cosine of 35 degrees multiplied by 30 m per second, multiplied by the time T. So in order to isolate and sol for tea, we're gonna divide, we have 75 m divided by cosine of 35 degrees multiplied by 30 m per second. And this gives us a time of 3. seconds. So we found the time it takes for the ball to get to the point that we're interested in 3.52 seconds. And now we can use that in our Y direction. 3.05, 2 seconds. All right. So now we can find this Y velocity. OK. We know V not Y we know a Y we know T, OK. That's three known values. One thing we're trying to find. So we're gonna choose the equation that doesn't include delta Y. We don't have information about delta Y and that's not what we're looking for. OK. So that equation, and again, we're finding the Y velocity in this step. And that equation is gonna be the following V F Y is equal to V. Not why plus A Y T. OK. So the final velocity that we're looking for is equal to the initial velocity, which is sine of 35° Multiplied by 30 m/s plus the acceleration negative 9.8 m per second squared multiplied by the time 3.52 seconds. If we plug this into our calculators, we're gonna get a final y velocity of negative 12. meters per second, huh? All right. So we found our X velocity, we found our Y velocity. And now we can use those to calculate the angle of the motion. And you'll know that the Y velocity is negative. That means the ball is going downwards. OK. So we're kind of on the second half of this trajectory, the ball has already reached its maximum and it's coming back down. So let's draw out what we have. So our X velocity is positive, we've chosen right to be positive. So the ball is moving to the right and it's also moving downwards because the Y velocity is negative. Yeah. So we have some motion in the right direction. OK? 24. m per second. Now, we also have motion in the Y direction pointing downwards 12.7023 m per second. Yeah. And what we want to calculate is the angle that this motion makes. OK. The angle between these. Now this angle is gonna be related through tangent. OK? Because we're talking about the opposite and adjacent sides. And so theta is going to be equal to the inverse tangent Of the opposite side which is the vertical side, 12.702, m/s divided by the adjacent side 24.5, 7 456 m/s. And this is gonna give an angle theta of 27.33°. Now this gives us the magnitude of the angle. OK? I use the magnitude of the velocities here. So we have just the speed. So I didn't include any signs inside of this tangent. OK. But if we look at our angle, it's going downwards from the horizontal. So that indicates a negative angle. And so our angle theta is actually going to be negative 27. degrees. If we go up to our answer choices, we can see that that corresponds with answer choice. A OK. The angle of the bowl is negative 27.3 degrees. Thanks everyone for watching. I hope this video helped see you in the next one.
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