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An arrow is shot at a stationary bird witha starting velocity of (20 m/s) with anangle of 65 degrees with respect tothe horizontall. 1a. What is the top height of t...

Question

An arrow is shot at a stationary bird witha starting velocity of (20 m/s) with anangle of 65 degrees with respect tothe horizontall. 1a. What is the top height of the arrow right before ithits its target? 1b. What is the vertical component ofthe arrow’s velocity, right as it reachesthe top of its motion and hits itstarget? 1c. What is the horizontal distance traveled bythe arrow, right as it reaches the top of itsmotion hitting its target?

An arrow is shot at a stationary bird with a starting velocity of (20 m/s) with an angle of 65 degrees with respect to the horizontall. 1a. What is the top height of the arrow right before it hits its target? 1b. What is the vertical component of the arrow’s velocity, right as it reaches the top of its motion and hits its target? 1c. What is the horizontal distance traveled by the arrow, right as it reaches the top of its motion hitting its target?



Answers

An arrow is shot from a height of 1.5 m toward a cliff of height H . It is shot with a velocity of 30 m/s at an angle of $60^{\circ}$ above the horizontal. It lands on the top edge of the cliff 4.0 s later. (a) What is the height of the cliff? (b) What is the maximum height reached by the arrow along its trajectory? (c)
What is the arrow's impact speed just before hitting the cliff?

This example will be doing a projectile problem where we shoot an arrow. The top allege that is 1.5 meters tall and we shoot it so that it sticks into the edge of a cliff of Heine H. Now, this whole process takes four seconds. We shoot our era with an initial magnitude of velocity of 30 meters per second and at an angle to the horizontal of 60 degrees. So the first thing we're gonna figure out is what the height of this cliff should be. After that, we're gonna determine what the maximum hide of our trajectory is. So that would be this distance here. And then finally, we're going to determine what the impact speed or the magnitude of velocity is when our aerial, it's the cliff. Okay, so for the first part, we want to determine what ages. So H is going to be our final Y position. So we want a cinematic equation that involves heights, times and velocities. And the one that comes to mind for me is the difference in height, it's equal. Teoh. You know why Times t minus 1/2 G square, where instead of just putting plus 1/2 a t squared. I went ahead and substituted a for minus g already. Since we know we're only dealing with the acceleration due to gravity. Okay, So in this case, why not is going to be equal to our starting height, which is 1.5 meters? Why is just going to be equal Teoh behind we end up at which is going to be age we not lie or why component of velocity is just gonna be a V. Not times the sign, uh, our angle Fada and then everything else. It's stuff we have. We have the acceleration due to gravity and we have our time, period uh, Delta t. So we'll substitute out to you for a Delta T. So let's put all that stuff in. We'll just leave. Why not? As it is for now and then we'll solve for h toe h minus. Why not equals v Non times the sign of data times Delta T minus 1/2 G Delta T squared. So two sulphur Rachel we have to do is just add why not to both sides. We'll just put it up here it easier, all right. And there's our expression for each. So now we just need to close all of this stuff in. We've got the not fate a Delta t G. And why not? So if we solve for age, we will end up with a height of 27 meters for the clips so you can see that my drawing is definitely not to scale. That's already. Okay, so for part two, we're gonna look at the maximum height that our era will reach. So usually the way I like to do this is I like to start at the cinematic equation. Uh, the final velocity and the Y squared when it's the initial velocity in the Y squared is equal to minus two g times the change in height. And then I remember that when we're at our maximum height are y velocity is momentarily zero so we can allow he finally go to zero, and then we can just substitute the rest of our knowns into this equation and solve for why? So we can call this why, Max, given the stuff that we've substituted in, so it looks like we gotta cancel ar minus signs right away. And let's just right, what we're left with Well, you know why Squared equals twice G and his lie Max minus. Why not? And now we're gonna sold for why, Max? And also substitute out. You know why? With our expression for the y component of our velocity, which is gonna be v non times the sign, Uh, 60 degrees, fourth data. Okay, so if we do that and divide by two g will have dina times the sign of beta divided by two g equals Y max minus Why not? And then we'll add why not to both sides. And we will have our final equation, which is that the maximum line is just equal to you Are starting height waas be not sign data founded by two G and actually, I forgot to square RV not why they're so we should actually have squares on wien on and sign of data. So 10. So usually this is just our expression for to the maximum height when we're just shooting on flat ground and it hits the ground. But since we're not on flat ground, we're starting at 1.5 meters. We have to add this additional height here. So if we plug in all of our knowns. We will get that our maximum height is 35.9 meters. Where are you? In? We'll get sit down. For the last part of our problem, we're going to find the impact. Speed is the final. So to get the final, we know it's just gonna be equal to you. Square root of the sum of the squares of its components. Right to a lot of the final X squared, plus the final y squared. Now, of course, the final in the X for projectiles on Earth is always just the initial X philosophy because it doesn't change during the flight. And of course, we know that the X component of velocity is just given by, ah, the magnitude of our initial velocity times the coastline of the angle. We fired it up, so that settles B not X. But for the final, why will have to do something a little extra? We will actually have to use an equation or one of our key thematic equations to solve for it. So we want an equation that gives us a final velocity. We know our initial velocity, and we know things like time as well in an easy equation to use that involves all that stuff is the final. It's gonna be equal to our initial velocity class the year acceleration times the time that it accelerates where in our case, our acceleration is just minus G and our initial Y velocity is just be not times the sign of data. All right, so if we plugged in our numbers for V not X over here, which was going to be 30 meters per second times because kind of 60 degrees, I believe that will give us 15 meters per second for view, not X. And if we use this expression over here for the final, why, we should get 13.3, specifically minus 13.3 meters per second. Right, since it's gonna be coming down when it hits the cliff like this, surely white component should be zero. Sorry should be negative. Okay, so now we can just go ahead and calculate our final impact speed. It's just going to be given by the square root of 15 meters per second squared plus a minus 13.3 meters per second squared, which will be 20 meters per second. All right, so there you have it. We determine what the height of our cliff must be given that we shot in a row Ladic from a height of 1.5 meters and it hit the edge in four seconds. We figured out what our maximum high is during our trajectory for arrow when we also figured out what the impact speed of our air awas.

Question 52 states that Archer shoots an arrow over a castle wall by launching it with an initial speed of 20 meters per second at 65 degrees above horizontal. Assumed the arrow lands on the other side of the castle wall at the same elevation as the launch point. Question A states. What is the maximum height? Does the arrow attain be what is a range of the arrows? Fight and see what is the vertical component of the arrows philosophy just before it lands the other side of the castle walls. So love the part to this question. First thing we should always do if we're gonna use X and y set up our Cartesian coordinate system and a diagram, although mostly scenarios is free. Simple, just drawn arc launch at some angle theta. And there you go diagrams air more useful when you're dealing with non uniform heights. So, like you shoot from a higher arrow to lower arrow. Sorry, A rose got arrows on the brain here. Projectiles launched from a higher point to a lower point or from vice versa, lower to higher. It's easy to recognize the situation. We draw diagrams, but always get practice to do it. Okay, so we want to know for part A what is the maximum height? So if you think they're projecting our maximum height occurs, I would see if it's a perfect protecting that which we always assumed. These scenarios fit occurs at a time of flight of alcohol capital t over to. So you know, when this occurs, remember our time of flight equation Capital T. I'll call here for time off late. It's to be not scientist over G. We want maximum height. So you want a time? I'll call little T to be the time of flight over to you. Put two in the bottom. So we have an expression for what time This occurs out, Max. The time at which the maximum height, the careers so de wise we're solving for here the distance in my direction at maximum height. I don't know. It got it sold for you. Well, look, we're given initial velocity angle and now time. It should be known that the where you can kind of figure out that the expression of the automatic equation excuse me to me to look at is our, uh, veena times t plus 1/2 a T squared. But remember, a white T squared so the distance is positive, going from zero to a higher point. So we saw a D Y initial velocity, and her white direction is positive as well. Let's be not science data time. We have an expression for it we could solve for, but it's easier. It's always better to leave things in terms of their variable values than playing numbers at the very end. And so now, acceleration due to gravity is negative, So even negative term here it's a negative 1/2. Uh, g times. Sorry. This is t squared above here. The line above, we have the not science squared. This whole time was just squared, all saying it'll be easier to deal with. Great. So if we expand this, we recognize one thing that the first term here is be not squared sine squared data over G. The second term is V squared sine squared data over G over to. So you have one minus 1/2. So this line is simplified. If you like to check on your own, please feel free. Um, but just as a heads up that this is what the expression is. This is a new reader. For each term we have a G in the first term at a to G and the denominator of the second term. So an expression does simplify it too, just to G and 11 fraction. So, looking at this, we have our initial velocity 20 meters per second. We have angle the angle, Fada. And of course, we know G. Sabina substituted our values and we find that the initial r se the maximum height of our projectile are a row here to two significant figures is simply 33 meters. Great party is what is the range of the arrow? Well, you know the expression for range given in our textbook. It's simply given us the velocities Initial velocity squared time sign of to theatre over G But if we look again, we know we not we know tha so we can solve For all these term itself, based on all our given terms again to two significant figures is just 61 meters in part, C finally asks what is the vertical component of the area's velocity just before it lands. So if we're given additional velocity in order to solve for the final velocity, we use the expression the not initial Why plus a whitey. We have our initial velocity wise. Just feed, not science data, which is positive acceleration in the wind direction is, of course, negative. G T. Well, one thing I should say. It's just before the air lands. So that occurs at a time of time of flight, which is capital t. So the time it takes is T, which is also just two times my little t from my expression that I drive the first point. So either which one? Whatever one. You want to use this accurate. We didn't solve for tea the time directly, but you could do that easily. So I'm gonna replace May little tea here with, um, two times. Well, I see this is a little bit, huh? This first lines a little bit inaccurate then. So I should say to be directly on us because it's the time of flight. This value here should be capital T. And he wants by here by two little T cereal because it's just before the air lands was the full course of the arrow. So it's just me not science data minus G. That was two V not signed. Data for G so we can find the thief velocity at the end point is simply negative. 25 meters per second. Negative, of course, because it's going to the negative direction. Yeah, There you go.

This question covers the concept of the projectile motion of a projectile which is launched horizontally. So for the projectile which is launched above the surface at high tech, the vertical displacement by is minus edge and from this weekend right minus edge is a cuBA into half of G times T squared. And from this the time of flight to music, you were into Square root of two times h upon G. So for part A the time taken to hit the ground, the secure and to square it off two and 2, The height that is 20 m upon The gravitational acceleration that is 10 m for a second. sq At the time T is secure in 2/2. Now, part B, the horizontal component of the speed at T equals one second, so at T equals one second. The horizontal component of the spirit is the initial speed that is eight m per second and the vertical component of the speed viva is minus off G times T and that is 10 m four second square into one seconds. Are we y equals -10 m/s. The speed are we is equivalent to square it off? Uh We x squared plus me way square or the speed is equivalent to square it off? Yeah. eight square plus 10 square. or the speed at the end of 1 2nd is equivalent to 12.8 m/s. Now, for part C uh The end of T equals two seconds the object hit the ground. So the vertical speed at equals two secondaries minus of G. That is 10 m four seconds square into the time T. That is two seconds. Are we wise -20 m/s? The speed is it covering two square out of v X square plus V vice square. Now that is the covering two square root off eight square plus 20 square are the spirits? Um, 21.5 m/s. The direction of the velocity bangle theater equals turning worse off the vertical component. That is very by upon the horizontal component. Or the angle theater made with hard rental is turning was off -20 upon eight. Or the angle theater, or the direction is minus 68.2 degree with 1000. So this is the velocity when the objected the ground.

Party in in unit Vector notation the first displacement Delta arse of one. This is gonna be equaling 2 60.0 kilometers. Our we're going to multiply this by 40 0.0 minutes divided by 60 minutes for every hour I had. And so this is giving us 40 0.0 kilometers I had. And so the second displacement we can say our sub too. Your mother dealt Delta arson too. This is gonna be equaling two 60.0 kilometers for our multiplied by 20 minutes. Divide by 60 minutes for every hour. And this is giving us 20.0 kilometers. And it's it's more appropriate to say the magnitude of thes second displacement because its direction is 40 degrees north of east. And so with we can say delta arse up to in unit vector notation would be 20.0 kilometers multiplied by co sign of 40 degrees. I had some of your ex component plus 20 0.0 kilometers multiplied by sine of 40 degrees J hat. And so we find that the second displacement in unit vector form is going to be 15.3 kilometers. I had plus 12 0.9 kilometers J hat. Uh, the third displacement then is equaling negative 60.0 kilometers per hour multiplied by 50 minutes, divided by 60 minutes per hour. And then this would be I hat and this would give us negative 50 0.0 kilometers I had. And so the total displacement we can simply say Delta are would be the sum of all the 1st 2nd and third displacements. And so this would be giving us 40.0 kilometers I hot plus 15.3 kilometers. I had plus 12.9 kilometers J hat minus 50 0.0 kilometers. All right. And so we can say that, then our total displacement in unit vector notation would be giving us 5.30 kilometers I hat plus 12.9 kilometers. Jay had the total time for the trip. Delta T would be simply 40 minutes plus 20 minutes plus 50 minutes. So 110 minutes and we can then say that this is approximately equivalent to 1.83 hours. And so we can then find for party the average velocity vector. This would be equaling 25.30 kilometers. I had a plus 12.9 kilometers J hat invited by 1.83 hours and this is giving us 2.90 kilometers per hour. I hat plus seven point 7.1 kilometers per hour J hat so that being unit vector notation, however, we want to find the magnitude of the average velocity and so this would be giving us the square root of 2.90 kilometers per hour. Quantity squared plus 7.1 kilometers per hour quantity squared and this is giving us 7.59 kilometers per hour. So this would be our magnitude of the average velocity for the entire trip and then for part B, we want to find the direction so the angle theta is given as arc tanne of the why component of the fun of the average velocity divided by the X component of the average velocity. And this is giving us arc 10 of 7.1 kilometers per hour, divided by 2.90 kilometers per hour, and this is giving us 67.5 degrees again we have a If we were to apply a Cartesian access to this problem. Ah, positive X value on a positive Y value will place us in the first, uh, first quadrant. And so we can say that the direction theta is gonna be 67 0.5 degrees u north of east. This would be our final answer for part B. That is the end of the solution. Thank you for watching.


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