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T-Mobile LTE3.34 PM100%XQuestionBuilding Unanswered0 = A projectile is launched atan angle 4 with respect to the horizontal The initial speed of the projectile is ...

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T-Mobile LTE3.34 PM100%XQuestionBuilding Unanswered0 = A projectile is launched atan angle 4 with respect to the horizontal The initial speed of the projectile is unknown building of height h-20.0 m is a distance L=50.0m from the launch position: The projectile just lands on the top of the building: What was the initial speed it was launched with?Respond with the correct number of significant figures in scientific notation (Use notation and only digit before decimaleg 2.5e5 for 2. x105)Type you

T-Mobile LTE 3.34 PM 100% X Question Building Unanswered 0 = A projectile is launched atan angle 4 with respect to the horizontal The initial speed of the projectile is unknown building of height h-20.0 m is a distance L=50.0m from the launch position: The projectile just lands on the top of the building: What was the initial speed it was launched with? Respond with the correct number of significant figures in scientific notation (Use notation and only digit before decimaleg 2.5e5 for 2. x105) Type your response Unanswered Submit



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A test rocket is fired vertically upward from a well. A catapult gives it an initial velocity of $80.0 \mathrm{~m} / \mathrm{s}$ at ground level. Subsequently, its engines fire and it accelerates upward at $4.00 \mathrm{~m} / \mathrm{s}^{2}$ until it reaches an altitude of $1000 \mathrm{~m}$. At that point its engines fail, and the rocket goes into free fall, with an acceleration of $-9.80 \mathrm{~m} / \mathrm{s}^{2}$ (a) How long is the rocket in motion above the ground? (b) What is its maximum altitude? (c) What is its velocity just before it collides with the Earth? (Hint: Consider the motion while the engine is operating separate from the free-fall motion.)

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Okay. Hello, everyone. Here it is, given a test. Rocket is fired vertically of birth from a well, A catapult gives it an initially speed 80.0 m per second at ground level, it's Indian. Then fire on accelerate a birth 4 m per second square until it reaches an altitude off 1000. We did At that point, Indian fields and rocket is going and to free fall. So after 100 m acceleration is deep. We have to calculate How long is the rocket in motion above the ground. Time for motion Above the ground. In second part beat What is its maximum altitude? By Max, We have to calculate what is experiences just before colliding with the earth. Final velocity at that time up. Collision Bill Ground, Let us to start solving it. Let ground level B do you know and point of one at the end. Up in German at the and off engine world, this is ground Jiro level. This is one level. Then it followed up. 500 turns back. This is turn. This is second maximum altitude 0.2 is highest point off altitude and third position just before impact with the ground for motion From 0 to 1 point we can write. We have a squared minus 80 square. It's called Took place off acceleration in tow. 300,000 m. So velocity at position one, we will get 1 20 m per second and time taken to reach the 0.1 would be we final. It's called Toby Initial plus n duty. This is 1 20. It's called toe 80 plus acceleration is four. So time taken toe attain the position one will be 10 seconds for position 1 to 2 Jiro minus 1 20 square. Initial velocity is 1 20. Finally zero acceleration is acceleration due to gravity 9.80 and distance final position upon initial position. So you will get final position minus initial position. Toby 7 35 m and the time taken to attain the third position. These final is called Toby Initial. Plus ain't duty. This is zero. Initial velocity is 1 20 minus 1.8 and two t. So time taken toe attain. The second position is 12.2 seconds. This is the time when it will attend the maximum height. Now for motion 2 to 3. Final velocity squared minus initial velocity is square, fellas. Equal to sorry toe final position minus initial position. Finally palace. It'd be required initially. Zero exploration is 9.8 and final minus. Initial position is 1000 plus 7. 35. That is 1735 m. So on solving it final velocity, that is, well, a city off impact with the ground we will get. Yeah, you will get 1 84 meter per second and time Bill B, we f is goingto be I plus 80. Final velocity is 1 84 initially zero 9.8 and to t so from here time, you will get 18.8 seconds. So First case, total time observing de time from 0 to 1. 12th one set to 12.2 seconds and 2 to 3. 18.8. So total time. Bilby 41.0 2nd. This is the answer off. Part eight. No maximum height report. Final position minus initial Toby 1000 plus 3 73. Sorry. Seven. 30. It becomes a 17 30 Meet it are 1.73 kilometer for C part. Yeah, Final velocity landing velocity will be 1 84 m per second in the downward direction. That's all for it. Thanks for watching it

So let's start answering any concept questions 1st. So for a question a it asked if the horizontal component of the initial velocity just being at X is proportional to the initial velocity. So we know the formula for horizontal component of initial velocity? Yes initial velocity multiplied by call sign of the initial. So as you can see from that equation then we can say that the horizontal component of initial velocity V. X. Is directly proportional to Vienna which is the initial velocity magnitude of initial velocity. Now question be asked if the time of like is directly proportional it's proportional to initial velocity. The formula for time like is this twice the initial velocity multiplied by sign of the angle over the acceleration due to gravity. As you can see from the formula we can say that the time of flight is directly proportionately initial philosophy. No questions see us if the horizontal range is proportional to the initial velocity R squared of the initial velocity given that Okay, it was stated That the ranges equals two. The product of the horizontal component of initial velocity in the kind of life. So from our equation from party Indy then we have Vienna causing data multiplied by to Vienna scientist and not over G. So simplifying then we have the nut squared two because I wanted to not scientists and not over G. And from the economic identity to casa and data. Mhm. Science data is equals to sine two. Theta. So our equation becomes like this. So as you can see the horizontal range is directly proportional to spread of the initial velocity. Now for the problem question. Mhm. It asked what if the range 623 m and now the initial velocity is doubled. So we have new initial velocity but the angle is the same. So taking up in that's not going to take a nap. What is the new range? So again, if R. S. Equals to Vienna Spread signed to take a nut over gee in his Echostar 23 m then are some any Sequels to? We're not and squared sine to beat up not in over G. So inserting the value of words we know, then we have to Vienna quantities grade modified by sign to data nuts. Is the under are the same over G. So simplifying. Mhm. Then we have four Vinod squared sine two Theta nut over G. But we know that This quantity here is just 23 m Equal to 20 of four times 23 m is equals to 92 m. As you can see, The new range is 92 m.

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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