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Vertical Motion Example 3
Hi, everyone!
Welcome back to integralcalc.com.
Today we're going to be doing another vertical motion problem.
And in this one, we've been given the position function y of t equals negative sixteen t squared plus ninety-six t plus fifty.
And we've been asked to find the maximum height that the ball reaches.
The position function is modeling a ball that's been thrown straight upward and we need to find the maximum height of the ball.
So the first couple of things we should do first of all is draw a diagram of this situation so that we can keep our thoughts organized.
It always helps to draw it out and identify exactly what we need to find.
So the ball is being thrown upward
and let's actually do this,
vertically upward, reaches a maximum height and then comes back down and hits the ground.
Now the reason, first of all, that I drew the position function starting from this height here
is because if we look at the formula that we have here for the position function,
it tells us a couple interesting things.
First of all that the gravitational constant is contained here and if your problem is in feet per second,
the gravitational constant is going to be thirty-two feet per second
and if you’re doing meters, it's nine point eight meters per second.
But this formula also tells us that the initial velocity is v sub zero here and that the initial height is y sub zero.
So assuming that your position function is in the form of this formula,
you can use these pieces of information in the formula to find for example v sub szero and y sub zero.
So as we can see, we have v sub zero here, ninety-six
and we have y sub zero here as fifty, which means, right, that if we identify t, y and v,
it means that the initial height of the ball as it's thrown upward is fifty feet
because this fifty here matches y sub zero and y as the height of the ball.
So the initial height is fifty.
The formula also tells us that the initial velocity v sub zero matches this ninety-six here
so the ball is thrown upward with an initial velocity of ninety-six feet per second.
So those are the two things we're able to gain from the formula.
We also know that time here is zero because that's when the ball is initially thrown up so that's time t.
So it's interesting because in the diagram, this is the point at which the ball is initially thrown upward.
This is the maximum height of the ball and then this is the point at which the ball hits the ground.
And at each of these points, we'll have time, height and velocity.
And we can use the position function and the velocity function to find all three pieces of information for each of these three points.
In this case, we only have been asked to find the maximum height the ball attains which is this piece of information right here.
The y coordinate at the maximum height.
So that's what we're looking to find.
And in order to do that, we need to first recognize that the velocity will be zero at this point
because remember, velocity is a combination of speed and direction.
So the ball is moving upward here; and then moving down over here.
So at it's very maximum height, at the very maximum point of its flight path,
it's stopped moving upward but it hasn't started moving back down yet;
which means that there's no change in direction and the velocity is zero.
So we know that the velocity is zero there.
And if we know that, then we can find the velocity function, set it equal to zero and use that information or our result to find y at this time.
So we need to know that the velocity function is the derivative of the position function.
The position function is what we've been given here, y of t, and the velocity function is its derivative.
So we need to find y prime of t which remember is the same thing as velocity, so v of t.
So finding the derivative, we'll get negative thirty-two t plus ninety-six.
That’s our derivative fucntion which is also our velocity function.
If we set that equal to zero, negative thirty-two t plus ninety-six, we can solve for t.
So we'll add thirty-two t to both sides, thirty-two t equals ninety-six and then dividing both sides by thirty-two, we get t equals three.
So this tells us that at time three, the ball reaches its maximum height.
But again, we really only care about the maximum height itself.
So we need to use the fact that t equals three and plug that into our equation for y of t, our original position function to find the maximum height.
So we'll find y of three...
and when we plug three in we'll get negative sixteen times three squared plus ninety-six times three plus fifty.
And when we simplify that, we'll get sixteen times nine which is going to be ninety and fifty four so that's negative one forty-four.
And then ninety-six times three would be two seventy so that's two eighty-eight plus fifty.
So negative one forty-four plus two eighty-eight is positive one forty-four plus fifty is one ninety four.
So one ninety-four is what we get here for y. That is therefore, the maximum height of the ball.
So our final answer is a hundred and ninety-four feet which is the maximum height that the ball attains during its flight.
and again you could use the position function and the velocity function to find any of these variables here if you needed to
because you know that y here is going to be zero.
Because when the ball hits the ground, its height is going to be zero
so you already know why its zero and then you could use the position function and the velocity function
to find the time at which it hits the ground and the velocity at which it hits the ground if you need to.
But that will be all for this video.
I hope it helped you, guys and I will see you in the next one.
Bye!