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The provided experimental data was collected by using motiontracking software to measure the vertical position of an objectover time. The initial launch speed of v0...

Question

The provided experimental data was collected by using motiontracking software to measure the vertical position of an objectover time. The initial launch speed of v0=12m/s13. Make a (properly formatted) graph of vertical positionvs time and fit it with an appropriate trendline. What is theequation of this trendline? (Use two decimal precision in yourequation.)14. Using the trendline equation, extract the initial launchangle of the projectile. (Round to the nearest 0.1)15. Calculate the overall ra

The provided experimental data was collected by using motion tracking software to measure the vertical position of an object over time. The initial launch speed of v0=12m/s 13. Make a (properly formatted) graph of vertical position vs time and fit it with an appropriate trendline. What is the equation of this trendline? (Use two decimal precision in your equation.) 14. Using the trendline equation, extract the initial launch angle of the projectile. (Round to the nearest 0.1) 15. Calculate the overall range of the projectile using the extracted angle. (Round to the nearest 1cm) Time (s) Vertical position (m) 0.05 0.484 0.1 0.943 0.15 1.337 0.2 1.787 0.25 2.258 0.3 2.558 0.35 2.698 0.4 3.147 0.45 3.362 0.5 3.578 0.55 4.08 0.6 4.008 0.65 4.188 0.7 4.343 0.75 4.752 0.8 4.583 0.85 4.667 0.9 5.02 0.95 4.813 1 4.679 1.05 4.815 1.1 4.731 1.15 4.673 1.2 4.496 1.25 4.623 1.3 4.444 1.35 4.461 1.4 4.272 1.45 3.743 1.5 3.708 1.55 3.59 1.6 3.234 1.65 2.808 1.7 2.637 1.75 2.093 1.8 1.838 1.85 1.328 1.9 0.979 1.95 0.532 2 0.041



Answers

Projectile flights in the following exercises are to be treated as ideal unless stated otherwise. All launch angles are assumed to be measured from the horizontal. All projectiles are assumed to be launched from the origin over a horizontal surface unless stated otherwise. For some exercises, a calculator may be helpful.
Firing from $\left(x_{0}, y_{0}\right)$ Derive the equations
$$\begin{array}{l}x=x_{0}+\left(v_{0} \cos \alpha\right) t \\y=y_{0}+\left(v_{0} \sin \alpha\right) t-\frac{1}{2} g t^{2}\end{array}$$
(see Equation (7) in the text) by solving the following initial value problem for a vector $\mathbf{r}$ in the plane.
Differential equation: $\frac{d^{2} \mathbf{r}}{d t^{2}}=-g \mathbf{j}$
Initial conditions: $\quad \mathbf{r}(0)=x_{0} \mathbf{i}+y_{0} \mathbf{j}$
$$\frac{d \mathbf{r}}{d t}(0)=\left(v_{0} \cos \alpha\right) \mathbf{i}+\left(v_{0} \sin \alpha\right) \mathbf{j}$$

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We are given the Parametric equations modeling the path of a projectile which is launched with an initial velocity of the not feet per second at a height of H feet above the ground at an angle of data with the horizontal. And we are asked to use a graphing utility to graph the paths. The projectile launched from ground level at different values of data and V not in each case were asked to use the graph to approximate the maximum height and the ring to a lot of parts. For this problem now, in Parametric, equations were given our X equals Vienna to co sign of data times T and why equals H plus v not signed data times T minus 16 times T squared in part A were given the data is equal to 60 degrees and that the knot is 88 ft per second. Now we are also told implicitly that the projectile is launched from ground level. This means that H is equal to zero and so are parametric. Equations become X equals V not which is 88 times the co sign of 60. Well, the co sign of 60 is the same as one half times t. So this is 40 40 and why is equal to zero plus V not, which is 88 times the sine of 60 degrees, which is root 3/2 times T minus 16 T squared, which simplifies to 44 3 t minus 16 t squared. And so I'll grab these two equations using Dez Moses Free online graphing calculator. So here is the graph for part A. Now, if you use the graph by zooming in, you can find the maximum height and the range of the projectile. These appear to be about 90.7 ft, and the range appears to be about 209.6 ft. Then in part B were given. The data is 60 degrees and Vienna is 1 32 feet per second. And so now the equations become X equals 88 times, one half again Times T or 40 40 and Y equals H zero plus 88 times. Route 3/2 times. T Sorry, not 88. This should be 132 times one half, which is 166 teeth. 9 166 but just 60 60 and this is 88 times Route three over to T. That's actually 1 30 to 10 through three of the two T minus 16 t squared, and this simplifies to 60 60 minus 16 T squared 66 times Route three T minus 16 to square. My mistake. To graph these equations, I'm going to use Dez Moses Free Online Graphing calculator. This is the graph of the path of the projectile until it hits the ground. If you zoom in on the graph, you'll find that the maximum height at the vertex of the parabola is approximately 204 to beat, and likewise, the max, or the range of the projectile, is approximately 471.6 ft. Then part C. We are told that data is now 45 degrees and vinegar goes back to 88 ft per second. Mm. So our equations become X equals 88 times the sine I'm sorry, the co sign of 45 degrees. This is route to over two times T or 40 route to see and why is equal to age, which is zero plus Vienna, which is 88 times the sine of 45 degrees or route to over two times T minus 16 t squared. This simplifies to 40 forward to times T minus 16 t squared. So we have two parametric equations. I'll plug both of these into gizmos. Here is the graph of the path of this tile. You zoom in on the graph. The maximum height appears to be approximately 60.5. Yes. Yeah, and the range of the projectile to where it hits the ground appears to be about 242.0. Finally, in Part IV, we're told that data is still 45 degrees, but now the initial velocity V not is 132 ft per cent. So our operations become X equals 132 times with 2/2 times T or 66 were two times teeth and y equals age, which is zero plus 132 times with 2/2 T minus 16 T square for 66. Route to T minus 16 T squared. Graph these equations using days Moses online permit. Yeah, and this is the graph of the path of the projectile. If you use the calculator, you can zoom in on the Vertex and you'll find at the maximum height of the projectile is approximately 272.0. What? Yeah, these and of the arranged points at which projectile hits the ground again, is approximately 544 points. Five Pete.

So we have this question. Ah, a spring gun at ground level. Fires right. A golf ball at an angle 45 degrees. The Boland's 10 meters away. Right? So once again, 10 meters away. That, he said, is respect to the origin. Right? So from here to here is like the ball lands here, right? This is an X component. The ball lands here and his 10 meters weight. Right. So this is this length Year is 10 meters And um So what is the balls? Initial speed. Right. So it fires it at an angle of 45 degrees and the ball landed here. Right? So what is the balls? Initial speed. Well, you have the range formula, right? The range formula contains initial speeds. That arrange formula is signed to Alfa Reggie Wright. Eso here? We did not know Jiggy. So we're gonna use the general one, which is 9.8 meters for a second right meters for a second square. That is the acceleration due to gravity. Ah, So what else do we have? What else we have? We have the angle, right? The angle is 45 degrees. So sign two times the angle What is the angle? 45 degrees times to let its 19 right side signed 90 degrees. Andi have to find this guy right here. Right? And what is arranged to range, which is this length here is 10 right, 10 meters. So this range here is 10 meters, right? So 10 is sign 90 degrees over 9.8 meters for a second squared. Right? So we're just gonna multiply, cross, multiply by So I cross multiply would be gonna get We're gonna get 98 right? Cause 10 times 9.8 is 98. And if this is gonna be us, what is signed 90 we'll sign off 90 degrees. Um is one right? So sign 90 degrees is one. So just have this. So you have a bee's Venus queer. The initial velocity squared as 98. So how do you find the velocity? Just take a positive square with Great, Because you don't think plus or minus because you don't have a negative velocity, right? Okay, we do have a negative velocity, but this time it this a positive velocity that we were looking for, right? Because it fire from the ground, right? It is it is accelerating it. It's not decelerating, right? Eso we have in the positive square root off 98. But on a tag later, his seven route to or approximately approximately mine point 899 So you could make it known for 90 year old right Could make it nine point 90 uh, meters for a second. So this is the initial velocity. Good. So to be part, what does it be porcine Perry asked to do in the be part if you've already saying that for the same initial speed sort of for the same initial speed that we found, uh, find it to firing angles that makes arrange six meters. Right. So we need to find two firing angles that make this range here, which is initiated 10. That's gonna make it six meters. Right, eso So this time the range should be six meters and we have to find at the same, uh, initial speed, which is 9.90 right meters for a second squared. Now you gotta finding angles so would not find an angle. And then what is the, uh, acceleration due to gravity? We're using them 0.8. So this time we're finding the angle. At first we new the range We knew the angle in the new tits and a region due to gravity. And have you found of initial velocity to be 9.90 meters for a second Now we knew they way. No, the, uh, acceleration due to gravity, We know the range, and I would know the initial velocity. Now we have to find the angle, right? So mechanicals, you know, cross multiply. So you cross multiply What is six times 9.989 point eight. That is actually 58.8. So 58.8. And what is number one? Nice squares. 9.9 squared eyes approximately. Uh, it's approximately 9.9. Square is approximately 98. Right. So, uh, this is gonna be, like 98 approximately and Ensign to Alfa right now. Divide bill side by 98. So, in a viable signed by 98 this is 58.8. They were having signed to Alfa to be 58 point eight over 98 8 That is 0.6. Then what is to Alfa to Alfa is gonna be inverse side. I said to Alfa is gonna be ever sign of 0.6. So it is in gross. I know. 0.6. In a sign of 0.6, it's approximately Ah, 38 point 36.87 So in verse and appoints is is approximately 38. 36 Vega pardon. 36.87 So there it is, 6.87 degrees. And then to find out if I just the viable side by two. So off finally is gonna be, um uh, so is 38 point 36.87 degrees or is 36.87 degrees or because it's not only 36.87 degrees. That gives you, uh, sign off sign Indra sign of, um 0.6 does not give you 30 6.87 It also gives you, Uh ah. 1 43.12 right 1 43.12 degrees. So all that it means is that, uh, you find signs of 36.87 degrees, you're gonna get 0.6. And if it's a fine sign off 1 31 43.12 degrees. You didn't get zero points is that is what we're trying to say, right? So that is what it means. So injuries sign off. 0.6 is 36.87 and 1 43 point I want to write. So if it divide both side by two, then you get in out far to be 18.4. So 18.4 degrees. That's when you divide by to or when Divide here by two as well. Then what are you gonna get? You're gonna get seven and 1.6 degrees. So you're getting out far to be 18.4 or 71.6.

Here. We're gonna do some equation work. So we're told that the range of a projectile is defined as the magnitude of its horizontal displacement. In other words, the range is the distance between the launch point and the impact point. On flat ground. Because we're working with flat ground, there's a couple consequences. Let's start by breaking this down. So we're saying that the project was launched with initial speed Visa vie an initial angle above the horizontal say to I'm going to try that right here. We have a vector of the velocity. Again with magnitude Visa by the angle above the horizontal. See that if we were to break this down into its components, we have the X. Where Visa of X is equal to the magnitude Visa by times the cosine of the angle and then we'd have the Visa boy where that is equally Visa I times the sine of the angle because it's opposite the angle. Since we're working with flat ground, the Visa Y initial which is visible item signed data and will be equal in magnitude to the supply final except they'll be opposite inside initially it leaves the ground like this and when it returns to the ground this factor in this victor the exact same in X. And why? But the wise are opposite in sign. We can back that up with our third equation by saying that V sub Y final squared is equal to v. Subway initial squared Plus two times a times of change and why? Well what is the change and why between the initial and final position they're at the same height So that change and why is zero? Which means this whole term goes to zero. In other words The square of the two is equal. In other words the magnitude of these F. Is equal to the magnitude evasive way initial and final. So now that we've broken that down, let's evaluate just how much time this will be in the air. So we can say that the amount of time spent in the air is equal to the change in velocity in the Y. All over the acceleration. And we know that acceleration to be negative delta G. That Visa boy final is going to be the negative of the Visa by initial. So it's going to be negative Visa by sign. Peter minus Visa by sign. Yeah, final negative initial minus initial. All over minus G. And of course this simplifies to two Visa by sign data. All over G. This is a change in time between the leaving the ground and the impact. So now let's ask ourselves what is the range? The range since Visa Becks is unchanging. If we multiply Visa backs by the amount of time we can find the horizontal distance and we define that horizontal distance to be the range. So to get that change in X, the magnitude of it, that range will say that that is equal to Visa of x times the change in time. And now we see why we've calculated the change in time using the values that we know. Mhm. First with substitute and Visa Becks which is a Visa by times the co sign of data and now accepted to our changing time which is to Visa by times the sine of data. All over gee. Let's begin to simplify this. This comes out to Visa by squared times two sine theta Hussein. Yeah all over G. And now we're going to use a trick identity. It is a fact that sign to sign data. Co sign data is equal to the sun. Two data. Mhm. Plugging this into our equation. We can say that the range is equal to V, initial squared times the sine, two times the angle all over G. And this holds true every time that we have a projectile that starts and land with no change in the white position on level ground. Thank you.


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