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(105) Froblem m below:eaple Mying nonzontallyAnred of 3.[ Ms when the fish in hcr talons wiggles loose and falls into theS09 Part (#) Calculate the magnilude of th...

Question

(105) Froblem m below:eaple Mying nonzontallyAnred of 3.[ Ms when the fish in hcr talons wiggles loose and falls into theS09 Part (#) Calculate the magnilude of the velocity o the fish relative the watcr #lxn hits thc waler 509 Part (b) Calculalc thc unglc. dcgtecs_ #hich thc fish' $ velocuy dircctcd bclow the horizontal whcn the fish hits the tuuct: tretrturanag Ieldinnt

(105) Froblem m below: eaple Mying nonzontally Anred of 3.[ Ms when the fish in hcr talons wiggles loose and falls into the S09 Part (#) Calculate the magnilude of the velocity o the fish relative the watcr #lxn hits thc waler 509 Part (b) Calculalc thc unglc. dcgtecs_ #hich thc fish' $ velocuy dircctcd bclow the horizontal whcn the fish hits the tuuct: tretrturanag Ieldinnt



Answers

An eagle is flying horizontally at a speed of 3.00 m/s when the fish in her talons wiggles loose and falls into the lake 5.00 m below. Calculate the velocity of the fish relative to the water when it hits the water.

In this example, we're going to determine what the final velocity is of a fish that gets dropped from in equals talents as it's flying over lake and it falls a distance of five meters before it impacts the water. So what we also know is that before the fish was dropped into the water, the equal is traveling with a horizontal speed of three meters per second, so we can go ahead and say that the fish was also initially traveling at a horizontal speed of three meters percent ing. So in order to figure out what the impact speed is going to be for the fish, we want to get the X and Y components of our final velocity vector. So if this is our final velocity that you're here, we want to figure out what the final why component of velocity is, and also what the final X component is, which we're going to get to first here. So the X component of velocity eyes not going to change throughout the entire problem because this is a projectile problem. And since gravity only acts downward and that the only accelerating force and the problem the X component will never be affected. So we could say that the initial velocity in the X direction is just always. Vieques never changes. Okay, so that settles one of our components for the y component, though we'll have to figure that one out from what we know. So ideally, we want to figure out what the final like opponent is given that we No, we start from a height of five meters and given that we also know but our angel, why component of velocity is zero since we only have a horizontal speed to begin with. And the equation that comes to mind that involves all of these values is the final squared minutes. V initial squared equals two times a chemistry change in height. Why minus? Why not? Okay, so in this situation we have that our initial velocity in the wider action is zero we have that are acceleration is just that due to gravity which will saves minus g and then our change in height is going to be why minus Why not? Which is going to be. Zero meters is where we end up. Minus are starting ICT, which is five meters. So why minus why not to be minus five meters. So if we ray all of that stuff down altogether, we'll have minus two G times minus five meters. So those minus signs will cancel out. And now we're just going to take the square root of the both sides of this equation, and we will get that the final velocity is equal to plus or minus 9.9 meters per second. So we'll have to decide whether we want positive one or the negative one. Ah, And in our case, we're gonna want the negative one because the down direction is negative in our recording assistant here. So I should say a recording system looks like this positive x this way was the boy this way. But ultimately, it's not gonna really matter what the sign is because we're just gonna compute the magnitude of our impact speed which won't care about the sign of our components. Okay, so now that we have our components of velocity for the impact speed, we know that the magnitude of our final velocity vector it's just going to be given by the square root of the sum of the square of each of the components so this will be the X squared plus the final in the wider Action square, where are X component of velocity again is three meters per second and our wind component is 9.9 meters per second. So if we go ahead and plug those values in and take the square root after sending him up, we will get 10.3 meters per second. So the final answer to our problem is that given that we drop this fish from a height of five meters into a lake with an initial horizontal speed of three meters per second, it will impact the water a speed of 10.3 meters per second.

Here we have another scenario, we say that an archerfish sees a meal sitting on a long stalk of grass To get it, it needs to successfully spit and strike the meal, but it's located 0.2 m away horizontally and 0.5 to five m above his mouth. So we want to know what is the minimum speed at which the archer fish must spit and what angle must it spit? So let's write down what we know. We know that the change in acts the horizontal distance from the fish is 0.2 m. And the change in why the vertical distance is 0.5 to five years. So the minimum speed with which the archer fishermen spit is the point where the top of the ark is perfectly horizontal. Of course the fish could spit at a much higher velocity, but we want to know the minimum velocity. What is the smallest speed that the archerfish can spit? And so at that point there will be no vertical component of its velocity. So let's think about it like this. We're going to have some Visa backs and some the supply such that the archerfish hits its target with a change in exit 0.2 and a change in Y a 0.5 to five. Let's start with the Y. So we know that change and why we do not know the Visa by initial, but we do know the Visa boy Final because we said there's no vertical component at the top. In other words, Visa boy Final is equal to zero to find the initial. We can use our third cinematic equation. Visa by final squared is equal to Visa Boy. Initial squared plus two times the acceleration times of change. And why we know that this value is going to be zero and we know that this value is going to be minus g minus 9.8 m per second square. And we know that this value is going to be 0.5 to five m rearranging. We can say that the supply initial is equal to they square it two times G times delta Y. Moving this part to the other side and taking the square rip plugging this all in. We say that the square root two times 9.8 times the changing y 0.5 to five gets us an initial velocity of 3.20 eight meters per second. Now we ask ourselves, how long was this in the air rather? How long did it take for it to reach the peak of its motion? To do that? We're gonna use our first cinematic equation and that's that the sub Y final is equal to the supply initial plus A tend to change in time, rearranging that we can say that the change in velocity divided by the acceleration is equal to the change in time. And we know that A. Is equal to minus G. So let's do this. The final velocity is zero. The initial velocity is 3.208 On the bottom we have minus 9.8 m per second squared and again the top is in meters per second. Doing this negative 3.208 And dividing by 9.8, we find that the change in time is zero point three 27 seconds. Now that we know how long it was in the air before it strikes the target, we can find the necessary beast effects. Would like to note that we have our abuse of Y. But we do not have our Visa banks are Visa Becks will remain constant and we know that it needs to travel this distance delta X. And this amount of time change in position over change in time will get us our visa Becks because it's constant. So a change in position 0.2 m that are changing time 0.3 to seven seconds. We'll get us our velocity. So let's do negative rather 0.2 divided by 0.3 to seven. And we get our Visa Becks to be 0.61 what meters per second. So now we have our X and Y components but we want to know what is the speed to do that. We need the magnitude and to do that. We're going to do the square root a Visa of X squared plus Visa boy squared, getting the magnitude of our vector with X and Y components using Pythagorean theorem. So again, that's the square root of 0.611 squared plus 3.208 squid. When we get that, we get a value of three point 27 meters per second. This is the minimum speed with which the archerfish must fit. 3.27 m per second. Let's ask yourself what is the angle with which it must fit. Let's draw out this triangle here, we have an X velocity of 0.611 We have a wide velocity of 3.2 year eight. And ask yourself what is this angle here? Well, that angle is going to be equal to the inverse tangent or are 10, whichever you prefer the opposite. Which in this case is 3.208 over the adjacent which is six or 0.611 plugging that in the inverse tangent of 3.208 divided by 0.611 We get a value of 79.2 degrees, 79.2 degrees above the horizontal. This is the minimum speed with which it must spit, and the angle above the horizontal that it must fit at the speed. Thank you.

Given the initial is food is 215 m per second and the angle of rejection is 50 to 52 degrees. The horizontal. So first of all, we will right. The vertical distance need to travel by the water is three centimeters or we can say this is a called a 0.3 m. Now we will use the kinetic equation as far as equal to u V i. P. Let's have game I p Square. So the vertical displacement He is equal to 0.3 m. Okay, equal to you by SV, not scientific. We noticed the vertical component of velocity, which is 212.15 and to sign off 52 Want to play with P? Let's have into a like that acceleration in the negative in the Y direction. It's minus 9.81 m for a second square. So this is the quality of minus my 0.81 p square so we can write. Get this equation when we call to 4.905 p. Squared minus 2.15 Sign off. 52 into T plus 0.3 is equal to zero. So from here we will solve this. So we have two values of time. Time is it called a 0.3 to 57 seconds and another one is T is equal to it was a quarter 0.0 1877 seconds. This is the second time. So these two times are correct. But here we need to take mhm. We need to take the time which so that target can reach in the least time. So we will take this value of time. It's a time when we call to 0.1877 seconds. This is the answer for part B. Now for part A. The audience understands covered is equal to X is equal to how digital component of velocity 2.15 to cost 50 to multiply with time which is 0.0 1877 seconds. So this whole gentle distance is a cool tool. 0.248 m. This is the answer for party. Thank you

Questioning any two references. Example for 14 page 16 So I encouraged you. Look at that problem first before going to this one just to get an idea of what we're looking for. But the Question 82 states that, as discussed in that previous example on Archerfish by dislodging unspecific insect from its resting place with a stream of water expelled from the fish's mouth. So in my diagram here, my fish Vivian, my origin. Down here, my fish shoots its water up at the insect, which is that black circle. So we supposed the Archerfish sports water with a speed of 2.15 meters per second at the angle of 52 degrees above the horizontal names for beetle on a leaf of a hate. The height was used in the previous example. It's always the same variables. Ah, hate of three centimeters above the water circus. And what horizontal distance from the Beetle should the Archerfish fire if it is to hit the target in the least time? So that's part a big questions will depart at first. Then we'll go to a party. Um, so we know that well, we're doing going how far. Wait, We're horizontal distance. It'll based on what we're given, we can look at the situation and say, Okay, well, that's very easily given by, you know, the English initial velocity of the extraction times time, which would be very easy to do if we knew what the time was. But we don't as of right now, so we need to stall for time, the time it takes for the water to reach that insect. So in most cases, if it doesn't work out for one of the directional equations, look at the other one, and usually that will give you the right answer. So do the same equation, including acceleration. Um, for the white component, we're going for me, the bottom to the top of this nearest upward. So my d white turns positive. My initial velocity is positive as well come shooting upward of the official shooting upward. What makes Hillary should do the gravity, of course, brings it down. Based on our coordinate system was 1/2 g t square. So, looking at this equation, I was gonna rearrange it to see if you have 1/2 g t squared. It's positive minus v not signed. data because the verdict component of our initial velocity is the sine of the angle corresponds to a side angle. Here, this is why minus the not silent e plus the wine which equals zero. Oh, not do I I'm not using the white very well. Sorry. I should've basis with h. Just what we'll do. Just one consistent with my variables here. So the height in which the insect is above the water is age, not d y. So, in the scenario, um, I want to solve her time. I know every other term of this equation, so that's a hint. Every need are quadratic formula. So by doing this, um, you get two solutions, so I use excel to compute this, use whatever format you use. Um, remember, but most cases we get two instances of time here. We have one solution to be 10.3 to 5 seconds, and sometimes we have a negative solution. But here you are a positive one so that both technically are viable the other time being 0.1 eight seconds. So you think about what's happening here. Why we have to. These times it means that at one height the height that we're looking at. Stay here. It occurs at the first time point. It means if it didn't have the insect, it would continue to travel upwards and go, then go back downwards that the other time would be something like this. Here. These are two points in time. Something like this may not be the skill, but that's the idea. Um, the question does say, however, if it doesn't hit the target in the least amount of time. So therefore we must need this time because a shorter So if Well, I guess here my question. I did represent the distance in the extraction simply by D, so I will keep that notation. So this distance is given by the not coast idea my horizontal component of my initial velocity times time which we now found to be this turn here. So sub seeing this value in for time at this point will allow me to calculate the distance that this water sorry that the beetle is in the horizontal component to me. 2.49 centimeters. The three senior figures, of course, 2.49 see him. So the point to 0.0.249 But that's going right back to centimeters, Whichever one you prefer. Both aren't technically correct. And be well, can already don't be at the state's How much time will it be? It'll have to react, but, well, it racks in the time it gets from the fish's mouth to the beetle itself. So that time we found in part a t peaceably 0.1 88 seconds again 236 figures. So a very cooked match time or else that beetle will probably be eaten for lunch.


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