>>Hi, this is Julie Harland, and I’m Your Math Gal. Please visit my website at yourmathgal.com, where you could search for any of my videos, organized by topic. In this video, we’re going to do a problem where someone throws a ball up in the air, and we know the initial speed of the ball. In this case, it’s going to be 48 feet per second. And we’re going to try to figure out how high the ball goes, before it starts coming down again. So, in this particular case, when you throw a ball up in the air, what happens is that this is the ground, and you — you’re standing on the ground, and you throw it up in the air. Actually, the height, you could sort of see, by this kind of a look, it ends up being a parabola. It takes the shape of a parabola. And this function, h of t, equals negative 16 t squared plus 48t, this tells you how high the ball is at any given time. Now, really, eventually, it’s going to hit the ground and stop. But if it kept on going, you could say, well, after, like, a hundred seconds, it would be, like, in the middle of the earth or, you know, through the other side of the earth, depending on how fast it’s going. But that’s what this function represents. It’s a parabola. So what we’re looking for is the maximum height. And if we look at this parabola, remember the maximum height is always the vertex point. I’m sorry. It’s — you could use the vertex point to get the maximum height, because that — that point — well, instead of x and y, we have t. So, the x-axis, in this case, we’re going to call it t, for time. And h is how high it is. So if we want to find out the height, we need to look at the vertex point of this, and we’re going to choose the second coordinate, which will be the height. Now, we can say, what would happen before you threw the ball? In other words, like, time is zero. No time has passed. Well, if you put in zero for t, into this equation, you’re going to get zero plus zero, or zero, and that make sense. It’s still on the ground. Then after 1 minute — I mean, I’m sorry — after 1 second, you would put in 1 for t, and you would get negative 16 plus 48. You’d say, well, it’s up at 32 feet, for instance, after 1 second. So the question is, we don’t know if that’s the height yet. We’re going to have to find the vertex point, and that’ll tell us our maximum height. So here’s the equation, if you throw a ball up at 48 feet per second from the ground, and we want to figure out the vertex point. So, to figure out the vertex point, figure out we — first we find out the t value, which, remember, is negative b over 2a. And in this case, a is negative 16 and b is 48. So we’ll just plug that in, putting in 48 for b and negative 16 for a. And we simplify that. You can see it’s going to be positive, because you have a negative and a negative, and you could do some canceling. I’m going to say 16 goes into 48 three times. The negatives cancel. So I end up getting three halves, or 1.5, if you want to write that in decimal form. It doesn’t matter which way you do it. So that’s going to be the first part of the vertex point. I’ll just put 1.5. So we know at 1.5 seconds, that’s when it’s going to start coming down again. That’s basically what it’s telling us about that ball. Now, how am I going to figure out how high it is at that 1.5 mark? Well, we’re just going to use that formula, h of t. So, I’m trying to figure out what happens when I put in 1.5 for t. You could use fractions or decimals. It really doesn’t matter. Let’s say you put in the decimal. 1.5 squared, plus 48, times 1.5. C.D. negative 16. 1.5 squared is 2.25. 48 times 1.5 is 72. And then we have to do negative 16 times 2.25. And that’s negative 36, I think. So we’ve got the h of 1.5 is 36. What does that mean? That means the height, at 1.5 seconds, is at 36 feet. So that’s the other part of the vertex point. And it’s asking us for the maximum height of the ball, and that would be this value right here, 36. So, we’ve got the maximum height of the ball is 36 feet. Now, if you remember, when we put in 1 for t, we got 32 feet. So it was almost there at 1, not quite. At a half a second more, it got to its top height of 36 feet. You could see what it is at 2 feet — I’m sorry — 2 seconds. You could plug in 2 for t, and you could see that it’s going to be shorter. So let’s just do that. What would it be at 2 seconds? You would say h of 2 is negative 16 times 2 squared, plus 48 times 2. So that’s negative 16 times 4. 48 times 2 is 96. Negative 64 plus 96. Think that’s 32? So, as you could see, it did start coming down at 2 seconds. So it seems reasonable that 36 feet probably is the correct answer here. And if you wanted to find out where it — when it hit the ground again, let’s say it asked that question. So the question is, when does it hit the ground? And we have the same equation we’re given. Keep in mind what we already know. At the beginning, we know at zero seconds it’s zero feet high. Right? And we now have that we figured out that at 1 second, it was 32 feet high. And we found the maximum, 1.5 seconds, it was 36 feet high. So then the question is, just out of curiosity, when does it hit the ground? So, at 1 and a half seconds, it’s up at 36 feet. I’d guess, well, it’s going to be back at the ground at 3 seconds. So we could check that out. So I wonder if that’s true. h of 3. What happens at h of 3? Well, we’d put in 3 for t. That’s the time. That’s negative 16 times 9, plus 144. 16 times 9 is 144. So we have negative 144 plus 144 is zero. So it makes sense, at 3 seconds, it would hit the ground again. We’re down here at zero feet. But that was my guess. Right? Right. How would you do that, if you didn’t know what the maximum height is. Somebody says, when did it hit the ground? Well, it hits the ground when the height is zero. Hitting the ground means the height is zero. Okay? So this height right here would have to be zero. So we put in zero equals negative 16 t squared plus 48t. And now we would have to solve this quadratic equation. There’s different ways of doing this. I’m going to do it by dividing both sides by negative 16, because I see that’s the greatest common factor. And then I’ll just have a nice t squared here. So, you could do it a different way, by factoring out 16 or factoring out a negative 16, but this is just another way. That gives me t squared minus 3t. And now to solve that, that’s just an easy quadratic equation to factor, t minus 3. So we get t equals zero or t equals 3, which makes sense, because we know at t equals zero it was zero feet high, and a second ago we found out that at t equals 3 it was also at zero feet high. So that’s when it hits the ground again. In any case, you could also check out what happens at 2 seconds. Right? I think we did that. Right? Already forgetting what we already did. Let’s see. Yes. We figured out at 2 seconds that it was 32. We figured out the highest it was, was at 36 feet. And so, that’s really the answer to this problem. The maximum height of the ball is at 36 feet, which happens to be after 1 and a half seconds. Please visit my website at yourmathgal.com, where you can view all of my videos, which are organized by topic.

@jklime : Let me know exactly where on the video you are referring to (in minutes and seconds) in order to answer you.

its relative to earth and physics its a constant in the formula

i have a problem like this, but they did not give me the speed. how do i solve it??? :

the height, h , in feet, that a certain arrow wil reach t seconds after being shot directly upward is given by te formula h = 112t – 16t(2) –>[the 2 in parenthesis is supposed to be squared so its actually t squared]

what is the maximum height of this arrow?

You have to find the vertex point of the parabola. Think of t as the x, and h as the y. The y-coordinate of the vertex point is the maximum height. It is the exact same type of question on do on this video where I do NOT give you t either. But I show you how to find the maximum height.

yeah i had the same question. i had to solve these types of equations in my algebra class, but i didn't really make sense to me because i thought they were using -16 in place of the gravitational constant in feet. but i thought it was supposed to be around 32 feet/second^2

1:00

That -16 is part of the formula. You would have to be given that formula to do the problem.

thanks 🙂

So, if you were to be asked how long does the ball stay in the air? what would you do to find that?

made not much sense but thx

This helped me! Thank you sooo much!

thanks!

Wow, these are some incredible mathematics skills. Definitely saved me. May the Lord have mercy on you.

Thank you math girl❤️❤️🤙

wth is 16t

You are the best, all the other videos provided the value of t within the question.

Thank you so much for this…. Made so much more sense then what I was taught.

this definitely beat my math teacher. thx :))))

Thankyou!

Thank you sooo much I got stuck in this on my math test and now I completely understand. I’ll be able to ace it tomorrow. You saved me. Thank you sooo much!!🤗

I love u <3

I hate this. I'm taking a class , otherwise I wouldn't care. But thanks for going nice and slow and explaining it well.

A ball is thrown upward from the surface of the earth with an initial velocity of 32 feet per second. What is the maxi-mum height that it reaches?

That's a win-win, the video came with a bonus such as when the ball hits the ground

THANK YOU! TOTALLY FORGOT THE FORMULA FOR FINDING THE VERTEX! Thanks!

whats the answer in this problem?

An object thrown verically upward with a velocity of 96m/sec. The distance S(t) above the ground after t seconds is given by the formula S(t)= 96t – 5t^2. How long it will take to reach the maximum height and how high it is?

Fortnit

No for jard pollls

thank you!!!!!!