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480 ft with an initial velocity of 16 ft/sec. How long does A rock is thrown straight down from height of ground? How long does it take before the rock Is ground? ...

Question

480 ft with an initial velocity of 16 ft/sec. How long does A rock is thrown straight down from height of ground? How long does it take before the rock Is ground? With what speed does it hit the it take to hit the traveled distance of '60 ft? 0f 112 ft/sec? When has the rock moving at speed its initial velocity if it returned to the Eround; What must have bcen Arocket Is shot stralght up (rom the earth in 20 seconds?Pogu

480 ft with an initial velocity of 16 ft/sec. How long does A rock is thrown straight down from height of ground? How long does it take before the rock Is ground? With what speed does it hit the it take to hit the traveled distance of '60 ft? 0f 112 ft/sec? When has the rock moving at speed its initial velocity if it returned to the Eround; What must have bcen Arocket Is shot stralght up (rom the earth in 20 seconds? Pogu



Answers

A rock is thrown down from the ledge of a mountain 200 feet above the ground with an initial velocity of 48 feet per second. If the height, $h,$ is given by the equation $h=-16 t^{2}-48 t+200,$ where $t$ is the time in seconds, how long does it take for the rock to hit the ground? Give your answer to the nearest one-hundredth of a second.

We have Lauren number 69. Ah, it is later toe the rocket propulsion question stays if your rocket is shot straight up in the air from the ground level with the initial velocity 240 the second East Side at. Okay, so model is given as equal toe 240 T minus 240. T minus 16 T square. Okay, this is our edge. Is hiding feet and T seconds after it launched. Okay, Now we have to find How long does it take for the rocket to reach a height of 800 ft? So we have to find the value off T van X equal to 800 ft. So we have just to plug in this value fate handed over here and solve this quadratic. So let us. All right. 800. Call toe 240. 80 minus 16. A square, A T square. Let us write it other way around at a 16 d. A squirrel minus 240 t plus 800 equal to zero. Now let us take 16 as common. We will be getting he squared minus 15 T plus 50 equal to zero. That has divide both sides. By 16, we will be getting d squared minus 15 T plus 50 equal to zero. Now we have to solve this. Quadratic. No, this is 50. So five into 10 and five plus 10 is already will become 50. So Victor will be the Esquire minus five D minus 10 T plus 50 equal to zero like a steak. T common from these two T minus five minus 10. Common from these two. T minus five equal to zero. So T minus five. T minus five can be taken as common T minus five. Two. Team I understand. Equal to zero. So, um, either this T minus five will become zero or T minus table become zero 30 minus 50 or T minus 10 is zero. So t equal to five and t equal toe 10 seconds. So there are two values off key. Let us check in in the original solution. Sorry. Originally question that if he satisfies the value or not. So if you plug into equal to five, it will become 1200 minus thesis 400. So 800 10, 2400, 608 100. Okay, so both are the possible solutions equal to five seconds and 10 seconds. Yeah. Let us, uh, learn what it signifies that if it is fired Yeah, from here. So after t seconds, if we re if it reaches equal to five seconds if it reaches here. So when returning, it will return after five seconds. So t equal toe 12th after it equal toe, 10 seconds will reach at the same height. So this is in going upward. This is equal to 10 will be returning downwards.

According to the question, we're given the equation for the height off the ball as Etch Off T is equal to re not. T minus 16 T Square, where a tuft is the height that the board reaches the knot is the initial velocity with which it is released, and T is the time in seconds were also given that the initial velocity with which the boys shot is 64 foot for a second, so we have to find the time in which it hits the ground. So when it hits the ground, hft will automatically become zero on me not is already given a 64 so this given equation will transform into let's number. This equation is one. So according to the question one will become zero is equal to 60 40 minus 16 T square, which implies that minus 16 t squared plus 60 40 is equal to zero, which implies that 16 T square, minus 60 40 will be equal to zero, which implies that if he takes extent common fact rising, so then we'll get T minus four is equal to zero. Now, from this step, what can we say? Which in place I'd owe 16 T is equal to zero or T minus four is equal to zero. So if 16 t is equal to zero than this implies that he's equal to zero. If t minus four is equal to zero than it implies, that is equal to four. Thought of zero and four we take. T is equal to four as the time. So what can we say? Therefore at a time, Deke will do four seconds. The bull hits the ground when it is released, with the initial velocity off 64 40%.

This problem. We have been given that height of a projectile is a projectile is launched from the ground level with an initial velocity of inner feet per second, neglecting air resistance. It citing 30 seconds after launch is given by S. Is equal to -16 T sq Plus. We not be in this question. We have been given that we're not is equal to 16 ft/s. No, we have to find the time that the projector will reach in the first part. Give any question we would be getting. It is equal to -16 days where plus 16 deep. So now proceeding further we will be obtaining an expression. The square is equal to -4. This is not possible and there would be no real roots to this. So therefore it is not possible for the hospital in this case to reach the height of 80 ft. Now in the second part, B we have been asked when it will be reaching the ground. So therefore the value of his food is equal to zero ft substituting the values of S. And we're not in the given the question, we would be obtaining The Question zero is equal to -16 T sq plus 16. Therefore, once we solve this we will be offering the value of T. is equal to 1/2. This is the answer. Thank you friends. I hope you like the video.

In this video, we're going to be dealing with derivatives and velocity and for those of you that need a quick refresher, A derivative is simply a modified function. F prime of X. That represents the rate of change at any point on an original function usually known as FX. And secondly, our velocity is just the derivative of the position function. Mhm. So in this question we are given that why is equal to 16 X squared and we know that, why is the distance fallen? And X is the time in seconds. So in this question they ask us when the ball hits the ground, if it has dropped from a height of 576 ft and to also find the velocity when it hits the ground. So we're going to break this up into two steps. We're first going to find when the ball hits the ground, then we're going to find the derivative of our function and then plug in our time that we're going to get from Step one. So part one. Let's find out where the ball hits the ground. So since we know that why represents our distance fallen, we know that our ball will hit the ground when it falls the full 576 ft. So if we plug in 576 for why we can then find when the ball hits the ground. So let's do that. We have 576, 0- 16 X. Squared. Now let's divide both sides by 16. And when we do that, we get that X squared is equal to 36. Yeah. Yeah. This leaves us with X is equal to the square root of 36. And we know that since our time cannot be negative, we're going to get positive six for our answer. So this means that the ball hits the ground at six seconds. So That was our part one for part two. Let's find our derivative two. So I'm going to use the first principles method for this. So The limit as h approaches zero of we know that when we plug in expose H we are going to get 16 X squared plus 32 H X plus 16. Mhm H squared. Then from this we're going to subject our initial function Which is simply 16 x squared. Please. Then we are going to divide all this by H. So if we simplify we are going to get Limit as H approaches 032 HX Plus 16. Each squared divided by H. We know that our issues are going to cancel out in the numerator and the denominator. Mhm. So what we're going to be left with is the limit. As h approaches zero 32 X plus 16 H. Mhm. Mhm. We know that at this point. If we want to find a limit, all we have to do is plug in zero for H. Leaving us with our final velocity function, velocity BMX is equal to 32 X. So we found our derivative. Now We are simply going to plug in six seconds into our velocity function. Now you can find out the velocity when the ball hits the ground. So our V six is going to be equal to 32 times six. And that's going to give us an answer of 192 meters per. Sorry, not meters. It's my Canadian kicking in 192 ft for a second. Mhm. And there we have it. We found out when the ball is going to hit the ground and the velocity when it hits the ground and there we have it


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