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If you throw a ball vertically up with velocity 19.6 m/s; what is the maximum height it will reach (assume it was thrown from height 0 m)...

Question

If you throw a ball vertically up with velocity 19.6 m/s; what is the maximum height it will reach (assume it was thrown from height 0 m)

If you throw a ball vertically up with velocity 19.6 m/s; what is the maximum height it will reach (assume it was thrown from height 0 m)



Answers

A ball is thrown upwards from the ground with a speed $\mathrm{u}$ for maximum range. (i) Find the maximum height reached by it. (ii) Find the angle of elevation at maximum height.

In this problem, we are given the speed with which an object is for upward is equal to 95 ft, fair skinned and the value of G is equal to 32.144 feet were scarce. Where we are asked to find the medicine height chaired by the object. As we know they it makes them hide french energy is equal to initial plan technology or we consider MGH is equal to half and we square where it is the maximum height and it is the initial speed. When we didn't this for H we get H is equal to we squared one by two G. So we will subside during his 95. Sweet spared divided by we first squared went by to manipulate 32.144 feet. Their skin square. When we saw this, we get H is equal to 140.4 ft. This is the amazing how to shoot by the object, which is the answer to this question. Thanks for what you would do.

So we're given this position function. And first what we're doing with this position function is we're determining what the height is of the ball at its highest point. So in order to do this, we need to find where the velocity is equal to zero. Because velocity is equal to zero right about at this point because the derivative is zero the the slope of the tangent line zero. So this is really what we're looking after. We need to figure out at what time does this occur? And then we can put that time into the position function to determine the actual height. So, um, if we find the velocity of this function, we see that it is negative 32 t plus v. Not with that, um, the velocity. Now we just have to find where it's equal to zero. That's gonna be once again negative 32 t plus we not we saw for tea and see that t is going to be equal Teoh v not over 32. And we get that because we bring the 32 tea over here and then we divide by the 32 to get the T isolated So for Part B. Um, what we're going to see is that before we actually get to part B, what we have to dio is we have to now place this time into our original position function. So we now see if we evaluated at this point s a V not over a 32. This is going to equal negative 16 times. He not over 32 and that is squared, plus the not times e not over 32. And at this point, it's just simplifying to determine what this in fact is. And this is giving us the height at its maximum point. So we have 32 squared done there, and then we can cancel that out with the 16 to an extent. So it's gonna give us a negative the not squared over 64 and then So that way we can combine like terms. We're going to make this, um, 64 as well. So it's gonna us 64. And, Lord, to do that we have to double the numerator, so we will get to the not squared. And now that they have the same denominator, we can now say that the maximum height of this position function will be vey not squared, divided by 64. So that's part a, um now moving down to part B into it in green. So for part B were being asked more specifically to figure out the velocity with which the ball strikes the ground. So this we need to know when the ball strikes the ground the time at which the ball strikes the ground and when something strikes the ground, it means it has a height of zero. So what we need to figure out is, when is this going to equal zero? Okay, so we set zero equal to negative 16 t squared, plus veena t. What we can do here is we can factor out of tea. At this point. It just becomes pretty basic algebra. So I will give us a negative 16 t plus B not. And here we now have it equaling zero. So we see two cases we see t equals zero. Um, but with t equals zero, this doesn't really tell us anything when the time is equal to zero. This is the right before the ball is actually launched. So obviously the ball is not at any height because it has not yet been launched. So t equals zero we don't really care about because has nothing to do with the ball actually striking the ground because the ball hasn't even been launched yet. What we do care about is when we have zero equals negative 16 t plus me not because now we can solve for T. When we add 16 t to both sides and divide by 16 we see that t is equal to V not over 16. We're not quite done yet, though, because we just found the time in which it occurs. But what we really need to find is the velocity, um, of the ball when it strikes the ground. So we want to refer back to our velocity function, which is right here. Um, so we're gonna have v of the not over 16 and we're going to put this in place of tea when we're looking at our velocity function. So it was negative 32 t. So it is in fact, going to be negative 32 times being not over 16. And then that plus close Amina, right? And once again, just come just simple computing tells us that's going to be a negative to the not plus Vina, and that will simplify too negative v Not so. That is the velocity with which the ball actually strikes the ground.

Uh So in this problem, um I'm under the assumption that you have established that the acceleration, the problem With gravity is negative 9.8 m/s squared because when they talk about the initial velocity being 19.6 meters per second With an initial height of zero uh then the equation that we're looking at is that negative 9.8? I'm going to call it X instead of t squared Plus 19.6 X is the equation that we're looking for a lot of people um maybe right as sfx, you know, for the height. So if you look at the question, how high is the ball thrown? Um What I would probably do is just factor out an X in the problem. So negative 9.8 x plus 19.6. Uh And what that will tell me is the time it will take for the ball to return to the ground. Because this one is X equals zero. Uh I guess I should point out that if effects he goes zero. Whether if if this is zero then the height of the ground. zero. So that negative 9.8 x plus 19.6 Equals zero would be when it returns. So what I would do Is subtract that 19.6 to the right side and then divide by that negative 9.8. So the time that it takes to return use a calculator because I don't know what that is in my head, 19.62 by about 9.8 Gives Me two seconds. So let's think about that for a second. That's actually your answer. Part B is a ball goes up back down from zero seconds to two seconds. Well at one second it's going to reach its maximum height. So what I would have to do is go back to the original problem and plug in one and from both of these, And they have 9.8 times one plus 19.6 times one would give me a maximum height 9.8 plus 19.6 gives me a maximum height for part A of 9.8 m, um and that was from plugging one in to this.

Okay, So in this problem we have a ball that is thrown vertically, upwards and when it is four meters above the ground. So me start recording this information when we are four meters above the ground, its velocity is half of what it started with. So this initial velocity is going to be twice as much as that final velocity. Okay. And we also know that we're on the planet Earth. So my acceleration is negative. 9.8 meters per second squared. Now, in this problem here, our goal is to figure out how high the ball will eventually reach if it is at half speed of four meters in the air. Um, which means we're actually gonna need to do this in two steps. Because in order to answer that question of how high will it get? We first need to know how fast was it actually going when it left the ground. So that means looking at my numbers here. Um, I have one unknown, and that's that velocity. Ah, because I have three numbers I'm looking with or working with, and the one variable we don't know and aren't looking for is time. So I'm gonna be using my Kinnah Matic equation over here on the right. That does not have anything to do with time. And that means my equation is V squared is equal to be not squared. Plus two a axe. When I plug, my number's in here. That final velocity of 1/2 of the becomes 1/4 because that half gets squared as well. V squared equals V squared plus a two times a negative 9.8 times of four gives us a negative 78.4. Now, when I tell you this help a bit since I have to be squared on both sides, I can combine my leg terms. And then since both sides are negative, we can straighten that up a little bit and we end up with 3/4 of that velocity, um, squared is equal to 78.4 and then I will multiply both sides by 4/3 to cancel that fraction. We have these squared equals 104.5 and take the square root to show that our final velocity is 10.2 meters per second. Now that again is the first part of the problem because this tells us how fast the ball is initially going. Our end goal here is to figure out how fast will it be going or not? How fast, How high up it will actually get when it comes to a complete stop. Um, to me dribble more of my tables in there. There we go. So we need to do one more of these problems and this one's gonna be really similar to what we just did, because the initial velocity is our answer from before. The initial velocity is 10.2 meters per second. The acceleration is still negative. 9.8 meters per second squared. Only peace that's different here is we don't know what exes. That's actually what we're trying to find. But we do know this ball, uh, rises to its maximum height, so its final velocity is going to be zero meters per second. We're looking for how high up it is when we reach zero meters per second. Again, we don't know time, nor do we care about time. So we're going to use that same equation. V squared equals V not squared, plus two a x. When I plug in my numbers. Here we have zero squared, which is zero is equal to the not squared. So 10.2 squared plus two a x. So multiply those together to get negative 19.6 x. Then I'm gonna add that over and take the school and square that number. So we get 19.6 x equals 104.4 Divide both sides by 19.6 and I get my final answer. The maximum height this thing will reaches 5.31 meters off the ground.


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