Height of an Object If an object is projected upward from an init...

Question

Height of an Object If an object is projected upward from an initial height of 100 ft with an initial velocity of 64 ft per sec, then its height in feet after t seconds is given by s(t) = -16t^2 + 64t + 100. Find the number of seconds it will take the object to reach its maximum height. What is this maximum height?

Accepted Answer

Hey, everyone in this problem, a ball is thrown upward from the roof of a building with an initial velocity of 16 ft per second. The height of the building is 50 ft. The equation for the height of the ball after two seconds is given S S F T is equal to negative 16 T squared plus 16 T plus 50. We're asked to determine the time when the ball will reach a maximum height and also to find the maximum height, we're given four answer choices, option A 0.5 seconds and 54 ft. Option B one second 50 fee, option C 0.8 seconds and 52. ft or option D two seconds and 18 ft. So if we imagine this ball being thrown from the roof of a building, OK. We're giving the equation of its height, which is a Parabola, OK? A quadratic equation. So we can imagine that this ball is gonna make some sort of parabola like this. It's gonna be thrown, it's gonna go upwards and then it's gonna fall back down. So when we think about finding the maximum height and the timer that occurs what we want to find is the vertex. Hm So if we're writing this question in terms of math, what this question is asking us to do is to find the vertex of the quadratic equation we were given now one way to find the vertex is to put this equation into vertex form. OK. We put this into vertex form. We'll be able to pull out that vertex from that equation. So we're called that vertex form is Y is equal to a multiplied by X minus H squared plus K or the coordinates H comma K are the vertex. OK. So let's start by writing up the equation we have, we have S F T is equal to negative T squared plus 16 T plus 50. Sure. So what we wanna do is we want to complete the square with the first two terms. In order to do that, we're gonna leave those two terms negative 16 T squared plus 16. Now, in order to have this as something squared, we're gonna need negative four. OK. We're gonna need minus four. You'll see that in a minute. Oh If we have minus four, we initially had plus 50 we need to also write plus 54. OK. Negative four plus 54 gives us that original 50 we had. So we're just splitting this constant term at the end up into a piece that's gonna go inside of our perfect square and one that's gonna stay outside Now, when you look at vertex form this perfect square, this X minus H square term, it is just an X. It doesn't have any coefficient in front of it. Other than one that coefficient is outside of the perfect square. So we want to factor this coefficient of negative 16 out first. So we're gonna factor negative 16 oh of the first three terms. So we have negative 16 multiplied by T squared minus T plus one quarter. And then we have our plus 54 term hanging at the end. Now T squared minus T plus one quarter, we can write this as T minus a half squared. And so our function becomes negative 16 multiplied by T minus one half squared. OK. So you can see why we added this negative four in why we broke up that constant term was so that we could write the first three terms in this form. OK? And then we have our plus 54 at the end. Now, if we can compare this to our vertex form, we have A is negative 16, H is negative a half or sorry, H is positive a half. You have to be careful here. The formula has a negative in there. So we have a negative H is a half and K is equal to 54. OK. So we have that one half, comma 54 is our vertex. Now we had a negative coefficient in front. So this para opened downwards, which is what we expected based off our, our drawing right, the ball was gonna go up first and then come back down. So vertex is at one half, 54 OK? One half represents the T value and represents s of T which is the height of the ball. So the maximum height occurs at T equals one half seconds and the maximum height is 54 ft. OK. This corresponds with answer choice A OK. The ball will achieve its maximum height at 0.5 seconds and that maximum height is 54 ft. That's it for this one. Thanks everyone for watching. See you in the next video.

Key Concepts

Quadratic Functions

Vertex of a Parabola

Maximum Height Calculation

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