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(iv) A small, spring-loaded toy dart gun is used to shoot a dart straight up in the air: The dart reaches a maximum height of 16 m above the dart gun: The dart is t...

Question

(iv) A small, spring-loaded toy dart gun is used to shoot a dart straight up in the air: The dart reaches a maximum height of 16 m above the dart gun: The dart is then shot a second time, but with the spring compressed 0.75 the distance it was before: What is now the maximum height reached by the dart? Ignore air resistance.12 m 9.0 m 8.0 m 4.0 m 3.0 mA block slides down a smooth ramp, starting from rest at a height h above the ground: When it reaches a point on the ramp that is a height h/3 abo

(iv) A small, spring-loaded toy dart gun is used to shoot a dart straight up in the air: The dart reaches a maximum height of 16 m above the dart gun: The dart is then shot a second time, but with the spring compressed 0.75 the distance it was before: What is now the maximum height reached by the dart? Ignore air resistance. 12 m 9.0 m 8.0 m 4.0 m 3.0 m A block slides down a smooth ramp, starting from rest at a height h above the ground: When it reaches a point on the ramp that is a height h/3 above the ground, what is its speed? (4gh3) (2gh/3) (gh/3)Y (gh)" (e) this question cannot be answered unless we know the mass of the block



Answers

A spring-gun projects a small rock from the ground with speed $v_{0}$ at an angle $\theta_{0}$ above the ground. You have been asked to determine $v_{0}$. From the way the spring-gun is constructed, you know that to a good approximation $v_{0}$ is independent of the launch angle. You go to a level, open field, select a launch angle, and measure the horizontal distance the rock travels. You use $g=9.80 \mathrm{~m} / \mathrm{s}^{2}$ and ignore the small height of the end of the spring-gun's barrel above the ground. Since your measurement includes some uncertainty in values measured for the launch angle and for the horizontal range, you repeat the measurement for several launch angles and obtain the results given in Fig. $3.76 .$ You ignore air resistance because there is no wind and the rock is small and heavy. (a) Select a way to represent the data well as a straight line. (b) Use the slope of the best straightline fit to your data from part (a) to calculate $v_{0}$. (c) When the launch angle is $36.9^{\circ},$ what maximum height above the ground does the rock reach?

In this problem. We have a plot of some data here. Uh And this is range versus angle. And were asked to do a best fit plot, A linear type of plot based on this data. So this is designed to use this formula here are equals the not squared over G. Sign to say to. But here they're just graphing our range versus angle Ceta. Uh and of course it doesn't make a graph. And we see we see a maximum range here at this is this is 45° right here in the middle, which is where we should have our maximum range. But of course this isn't a line. Well we can make this into a line by instead of graphing are versus data, we graph are versus sign tooth data. So we're going to make a different sort of a graph here, that's like going to be this are versus sign to tha tha and that should give us something that gives us a bunch of points that seem to be in something pretty close to a line. And then the slope of this line here, mm will be v not squared over G. This value right here will be f. So we can just plot that. We just need to calculate sine two theta of all these different uh values of fada down here and it looks like they're in five degree increments. So if we got 2025 30 35 so on and so forth up to 70 degrees. And then these ranges we just have to sort of make a good enough, good enough guess like this first one here, this looks like it's and Like 6.9 is very close to seven and this one looks like it's 8.6, you know, just a little above 8.5. And this one looks like it's uh 9.6, a little bit above 9.5. You know, there's you, when you do this, you might have some slight variations, everyone's answers might not be exactly the same, but you have to just make some some reasonable predictions. So I went ahead and plotted this data here. And so we have here are X coordinates sign two times these values of data, 2025 down up to 70. And here's the values of the range that I plotted and these values just correspond to these values right here. And I plotted those, made a chart here. And if you put them into a graph, there are these blue lines and then you can just run a linear regression or best fit. And it makes this line. Now, if you don't have, I just use this using demos which is an online graphing calculator. You can also do it on a. T. I calculator. But if you don't have any of that, you can just make your best that line. Just draw it in and then you're just picking points along the line to calculate the slope. Now there's some great software in Dez most. And so it just calculates the slope right here for us. So that's really convenient. We just get that the slope m is 10.9 about But if you didn't know that and you were unsure, you could just look at the graph here and say, oh well it looks like it goes through 00 here. It's actually not quite as we see right here, but this is, We could just, we didn't know that. We could say it's 00 and then uh it looks like this, here's a point right here. Um that we could try and look at that seems fairly easy to determine. And it looks like this point right here would be a 0.45 comma five. And so if we wanted to calculate that slope, it would be easy because it's just the ratio of these two since this is 00. So if we weren't sure, we could say, okay, m equals about five Over 045. And if you make that, you're going to get approximately 11.1, I believe. 11.1. So that is pretty close To this value 10.9. So we see that we've got a pretty good good approximation here using this uh picking points along the line. The important thing is that the points we pick our along the best fit line. Not necessarily these data points we have here in blue, not these particular points. The points along the line. And that line may not include any of these actual particular points. We're looking for these we're going to pick points for measuring our slope that are actually on this best fit line. I'm gonna go ahead and use this 10.9 value we have here. And if you remembered we said mm equals V not squared over G, which equals about 10 0.9 four. Uh and that's going to be meters are going to be the units. And so we can then say, well v not it's just going to equal 10 0.94 G leaders square root And we're just gonna use 9.8 Fergie. And then that's meters squared per second squared when you square root, it's going to be meters per second. So that seems correct. So 10.94 times 9.8. Okay. And then we're going to take which is 107.2. And then we just take the square root of that. And that's 10.35. So the not Is approximately 10 3 five meters per second. So that is our launching velocity. And then we have a follow up question here which is asking us To find the maximum height above the ground at 36.9° launch. So we're saying that our launching angle Is 36.9° and we want to know what is the max height? Well to do this, we can just use the formula V squared minus V initial squared equals two G delta Y. And when it gets to its max height, Of course the square to zero. So then we can just say delta Y equals negative V, not squared over two G. And we should be careful that this this is really in the Y direction. So that's going to be negative V not sign data over two G squared. So that is negative 10.35 times sine. Uh huh. 36.9°. And this whole thing is going to be squared Over two times 9.8 and this will be in units of meters. And then all we gotta do is calculate that and there's a there's a negative down here. This would be negative 9.8. So these are going to cancel as they should and we're going to get a positive height. So yeah, you just have to put that into a calculator and Mhm. Mhm. Mhm. Right. And putting all this in and calculating, we get delta Y Equals about 1.9 seven meters. Mhm

Okay, So our question says that it wants us to try something at a sporting event. And it says show that the maximum height H attained by an object projected into the air said to the baseball, football or soccer ball is approximately given by H is approximately 1.2 t squared meters which I have written here where t is the total time of flight for the object in seconds. Assume that the object returns to the same level as from that which was launched. Doesn't think you're 42. For example, if your maximum height attained was h equals 1.2 times 10 to the fifth meter square, 1.2 times five squared is equal to 30 meters. The beauty of this relation is that age can be determined without knowledge of the V not or the angle fate or not. Okay, so in order to do this, I'm going Tio, consider downward vertical component of the motion which will occur in half the total time, okay, and then also take. And that's because the time it takes for it to go up and then back down is the time it takes for it to get to the very, very top at the very peak and at the peak, the total time it is going to take for a default. I'm sorry. The time is going to take for it to fall back Down from the peak is half of the total time. Okay, which I can go ahead and write that here. So the time for it to fall down equals 1/2 the total time. Which one? Tisa gap. Multi for total. Okay. And then the equation that we're going to use is well, why? This is equation to Dutch 12 b. Why is equal to why not the initial y plus Vina in the wider action? What times time? Because of you, Meters per second, We wanted to be in meters are solution of meters. So it's times, time, times 1/2 the acceleration in the white direction, right times t squared acceleration times t squared will give us meters, which is about, uh, the units we're looking for. Okay. And this is the total time. All right, well, let's go ahead in. Ah, that's not the total time. That's just the time that we're considering in the generic equation. So now Let's go ahead and plug in what we have for our situation. Well, what? We're trying to figure out his H. Why not? Well, why not? What is what we're considering here To be the time it takes to fall from the peak. So are why not Here is going to be zero. Right? Okay, we're going to consider that to be zero v Not why we're going to consider the velocity initial velocity to be zeros that's going to be plus zero. And then this is going to be plus 1/2 a of lie. Well, we're dealing with gravity. Here's that's going to gravity and then t squared that we're going to be using the time it takes for it to fall downwards. So that's 1/2 the total time. Okay, so this is 1/2 the total time, okay? And that is squared. Yeah. So now if we go ahead and continue on here, 1/2 squared is 1/4. 1/4 time's 1/2 which is out front of the G is 1/8. So this is 1/8 G is 9.8 meters per second squared since its acceleration. Right. And then we have the total time here squared. Yes, this is approximately. If you take 9.8 divided by eight. That's approximately 1.2 Mrs T squared units is meters. So therefore, box, I'd anyway, in approving the solution.

So we got a firework that launches upward and then explodes in all directions, all with the same philosophy. Well, we want to figure out what the minimum angle that's these projectiles could potentially hit the ground with their final velocity. What's the minimum, man? And we know that that minimum in what has to be from some projectile that would be fired perfectly horizontal. So we're gonna said that perfectly horizontal velocity set that equal to our velocity because I won't change in any projectile motion problems. It's ordinary resistance, and we're gonna have our initial velocity in the Y direction equal to zero, since there is no velocity in that direction. Initially, all of it is horizontal, and we want to go ahead and figure out the final velocity. Go ahead and use this equation here. So V why final squared is equal to be. Why initial swear plus two times acceleration the Y direction times the height, that the object falls, sandwiches, heights each well And since there's no initial velocity, this goes away. We swear both sides and we get a B. Why final equal to to G each putting in G. The acceleration due to gravity in place of the acceleration and the X component of the final velocity, you know, is going to be the same as which we had here. This will also be there X final and we could go ahead and find that angle by using our candid trick the inverse tangents of the Y component over the X component. So it will be the two GH the root of a through G H divided by the V that we get here. That initial velocity the and this will be our final expression for our minimum angle. This is the minimum angle you'll have in a situation like this.

Okay we're launching a projectile from the ground and we want to figure out the maximum large speed. So um velocity at the peak is going to be velocity initial and since this is vertical it's just velocity initial um minus G. T. Although I really should be using V p squared equals V zero squared minus to A. And then h so h since the peak velocity is going to be zero is going to equal v zero squared over to a. Which is G two G. So I didn't need this that B. Okay so what if it shoots it at twice that speed? So the speed was goodness, It was the square root of two G. H. Why is it having so much trouble writing V equals the square root of two G. H. So now uh V is gonna be too square roots of two G. H. Okay what angle should you launch the projectile? Okay so I think we still wanted to achieve the same height so um uh V Y direction at the peak and uh yeah squared is going to equal V one um in the Y direction. So that would be sine of theta squared -2 G. H. So uh V one squared sine squared of theta is going to equal to G. H. So tha tha will be the inverse sine of two G. H Over V one squared. That's not exactly right, There's got to be a square root in here somewhere um to G H overview one square then you take the square root of that and then you get the inverse sine Okay so that would be um I guess I can simplify that a little bit more as the inverse sine Of the Square Root of two G. H. Over V However V0 was The Square Root of two D. H. So that would be V0 over v. one But V one, wait a minute. V one is 2 V zero. So and V zero was a square to to G. H. So this would be This is two square root of two G. H. So it's really just one half. So the inverse sine of 1/2? Right, okay. Will be 30°. Okay. Yeah. How far will it land? How far away will land? Okay so now we got to do X. Direction um Ax equals V. Times the co sign of tha tha times T. But we also know that V. Y. At the peak zero is going to equal V one sign data minus G. T. So t. is going to be V one sign tha tha Over G. Putting those together x equals v one Cosign Theta Times T. Which is v. one signed data over G. So it's going to be V one squared over G. Sign data casa and tha tha but V one is two square roots Of two G. H. So two squared is four square to to G. H squared is two G. H. Simplifying this little bit more to Cynthia to cosign theta is the sign of tooth data. Um And that's just going to be four h. So it's gonna land at four H. Sign squared of data. But the science squared of data sign because the sine of theta is one half So 1/2 squared is 1/4. So it's just gonna land at H. Interesting. Very interesting. All right. Thank you for watching.


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