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Part A: you shoot an object straight up into the air with anunknown velocity. The ball reaches a peak of 2m. what is theinitial velocity?part B: you shoot an object...

Question

Part A: you shoot an object straight up into the air with anunknown velocity. The ball reaches a peak of 2m. what is theinitial velocity?part B: you shoot an object straight up into the air with anunknown velocity. the ball lands 2s later (at the same height asthe launch). what is the initial velocity?please no cursive writing. sometimes its hard to understand! iprefer it in print when showing work so i can understand what i amlooking at!

part A: you shoot an object straight up into the air with an unknown velocity. The ball reaches a peak of 2m. what is the initial velocity? part B: you shoot an object straight up into the air with an unknown velocity. the ball lands 2s later (at the same height as the launch). what is the initial velocity? please no cursive writing. sometimes its hard to understand! i prefer it in print when showing work so i can understand what i am looking at!



Answers

A ball is thrown straight upward. At 4.00 $\mathrm{m}$ above its launch point, the ball's speed is one-half its launch speed. What maximum height above its launch point does the ball attain?

For a ball which is being thrown initially vertical direction. The velocity has given us the velocity squarish we share with the city square one is to key into the high priest wife. And when the ball reaches maximum the fundamental city this becomes zero. And from this the height. Why is the Cuban the squad of the initial velocity one said, this is a question at the Midway. Uh, by Midway will be the maximum height. That is we not square upon to g upon to so into one upon to. So we can say while Midway's we north square upon for cheap. That's a that's a question again. We can write the question can dramatic question as the velocity squared is a cure in the initial velocity squared -2. Key into uh, height that is violent. So we can replace five m. We square upon food, cheap of the square of the velocity at the middle ways. We're not square upon two. And if you take the square out on both sides, we find the velocity at the mid ways uh, And 0.71 times the initial launching school

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

For this problem we're told that a ball is thrown upward and its height H one m above the ground after T. Seconds is given by H. Of T equals negative five T squared plus 25 T. Where T. Is greater than or equal to zero. For part A. We want to calculate the balls initial velocity so that means that we want to figure out what H. Prime of zero is to do that. We'll first need to figure out what H prime of T. Is. So it'll be negative 10 plus 25. So H prime of zero will be just 25 part B. We want to calculate its maximum height. So that's going to be figuring out H. Prime are figuring out T. T. Not such that H prime of T. Not equal zero and then calculating H. At T. Not so to do that, you look at our derivative. So we need 25 to equal 10 T. Or T. Two equal 25 or 10. Which I'll leave it in that form for now. Part I believe. Can that be? Yeah. That can be reduced down though because 10 is going to be five times to 25 is five times five. So we can reduce that down to 5/2. Part C. Oh actually we're not done yet. We need to do each of 5/2. So we equal negative 10 times five over to. That's 25 negative 10 times five is going to be negative 50. And then who? One moment? Oh I was plugging that back into the equation for the derivative. Whoops. Uh So H of 5/2 would be negative five times 5/2 squared plus. 25 times 5/2. Which we can write as being negative five cubed over four plus. So 25 is five squared so would be negative five cubed over four plus five cubed over to or 25 cubed two times five cubed over four minus five cubed over four. So it'll be five cubed over four and 25 times five would be 100 and 25 125 over four or so. That might be reducible. Do not believe so for part C. Were asked when does the ball strike the ground? And what is the its velocity at this time? So we just need to sell the first of all, negative five T squared plus 25 T equals zero to get when it will hit the ground. So we can factor out a. T. So give us a negative five T. We can actually factor out of five. T. Will give us a negative T. Plus five equals zero. Which means that if negative T. Plus five equals zero then that means T. Will have to equal five. Okay. And so now we just need to plug that into our first derivative equation. So S. Prime five, it's going to equal our first review is negative 10 T plus 25. So that would be negative 50 plus 25 or just negative 25. So it's heading towards the ground at 25 m per second.

The first part off the question saying is that we not equals 1 30 p for a second on its north zero as we're throwing it from ground level on expected Esti, that's the height is to be between 55 feet on. You find us using a calculator by the given. The creation that we're gonna take now has a maximum of 3 55 or not or desert that would lead start rienda equation, that is my 16 x quick. That's 1 30 x under and up on the craft burden to get the craft. Now we're going to find the maximum using the talc option. And there we have maximum function. I know we select the balance. I see that the maximum as Steve will be 1.5 extra five for the creation, which means that never reaches the hide off 3 55 feet. So we made a note of death that the Mac side it will ever reach a C 51.625 feet. Hence, the ball does not every beat 3 55 feet for, but we it sees that now we beat party. Does that the initial height as 30 feet on initial velocity is to 50 feet per second. We blood this and do our equation, which is basically our original equation. Lester defeat Not It's not on Retained won 52 to 50 under and of I Do Eyes 3 55 We see that clearly. Teachers, now we find the dying Admit agreed on the height. So we find the intersection point on the left. It's and then defect function. Get the value as 1.41 0694 Therefore, yes, the Mac side for the escalation is 1006.56 Too afraid. And yes, the ball reaches the height 3 55 feet. I'm a time one point for 31 069 Food that 1.5 seconds


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