Hi, everyone.
Welcome back to integralcalc.com.
Today we’re going to be doing a vertical motion problem.
And in this one, we’ve been given this word problem which says, ‘You throw a ball straight
upward from the ground with an initial velocity of 96 ft/s.’
And then we’ve been asked to find out how high the ball rises and how long it stays
aloft.
We’ve also been given this formula for the velocity function which is v(t) = at + v sub
0.
And there’s several parts to this formula but the first one is a which is acceleration.
And for this particular problem, we’ve been told to consider that a = -32 which is a constant
gravitational force.
Because the ball’s being thrown up from the ground, and it’s still fairly close
to the ground, we’re negative any kind of change in the force that gravity has on the
ball.
We’re just going to call it a constant -32 and that’s really common in these types
of problems.
t is our variable and we’ll go ahead and leave that alone and then v sub zero is the
initial velocity which is the velocity of the ball when it’s first thrown from the
ground.
And we’ve been told from the problem that that is equal to 96 ft/s.
So the first thing we’ll do is write an equation using this formula for the velocity
function v(t).
So we’ll get v(t) = at + v0.
We’re going to plug -32 for a, we’re going to leave t as the variable nd we’re going
to be plugging in 96 for the initial velocity since we’ve been given that in the original
problem.
And just to give us kind of a visual idea of what this looks like, here’s the ground
and we’ve got a ball that’s thrown straight up and eventually it’s going to reach its
maximum point and come back down and hit the ground.
So at each of these points, there are three values.
We’ve got the initial point where it’s thrown up from the ground, we’ve got the
point at which it reaches it’s maximum height, and then we have the point at which it hits
the ground.
Those are the three points that we’re interested in.
And at each of those three, we know that the ball has a certain height y.
It has a certain time t and it has velocity v. So there’s some things that we can fill
in here.
we know that the initial velocity is 96 feet.
So we can fill in an initial velocity of 96 feet.
The initial time t is always going to be 0 unless they tell you otherwise but that’s
kind of the start of time in this problem.
We only care about the flight of the ball so when the flight path of the ball starts,
time is equal to zero.
And we also know that the height is equal to zero because the ball is thrown upward
from the ground.
So we’re calling the ground a height of zero.
So the height there is also going to be zero.
The height is also going to be zero when the ball hits the ground which makes sense.
So those are some things we know right away.
We know that we’ve been asked to fin how high the ball rises so we’re going to need
the height of the ball which is going to be this y value here because y represents height.
And that’s the maximum height of the ball associated with this point right here.
and then we’ve also been asked to find how long the ball remains aloft.
So that means that we need to find this value here, time t because if we find that, we know
how long it’s been or what time is when the ball hits the ground and that will encompass
the entire time span that the ball is in the air.
So those are the two things that we need to find.
This path here, this visual that I’ve drawn is represented by the function y(t).
y(t) is the position function.
With these kinds of problems, though often talked about like a particle that moves along
the x axis, if it’s moving along the x axis, the position function is modeled by x(t) because
the change in position is horizontally.
It’s along the x axis.
But when we have a problem like this, vertical motion, and the ball’s moving up or down
in this case both, up and then down, the position function is modeled by y(t).
So in our case, the position function is going to be modeled by y(t) and that’s the function
we need to find in order to solve for the rest of the questions here in our problem.
So remember that with these kinds of problems, either x(t) or y(t), in this case y(t) is
the position function; and velocity v(t) is the derivative of the position function.
If you wanted to know acceleration is the derivative of the velocity function but basically,
this is the order that we’re going in.
we have up here the velocity function so we’ve already got this one.
in order to find the position function, instead of taking the derivative and going in this
direction, we need to take the integral to go in this direction and find the position
function.
So, we’re going to go ahead and say y(t) is equal to the integral of the velocity function
which since we already know the velocity function, is going to be equal to the integral of negative
32t plus 96.
So we just need to take the integral of this to find the position function.
So taking the integral of -32t + 96, we’ll get -16t^2 + 96t and then we have to add c
to account for the constant of integration.
Now, to solve for c, we’re going to need an initial condition that involves this position
function y.
Well, what is the initial position of the ball?
We know that given this information here since it’s thrown straight up from the ground,
we know that y(0) in other words the height at time 0 is equal to 0.
So we know that y of zero is zero.
That means we can plug in zero for t and we can plug in zero for y of t and solve for
c.
So we’ll get zero equals -16(0^2) + 96(0) + c.
And obviously, we’ll get zero and we’ll see that c is in fact equal to zero, which
means that our position function y(t) will be equal to -16t^2 + 96t.
So that’s our position function.
Now that we have that, we can easily find y at the maximum height and t when the ball
hits the ground.
So in order to find the maximum height of the ball, we need to maximize this function
here.
and the way that we do that is by taking the derivative, setting it equal to zero and solving
for t.
So we can take the derivative, we get -32t + 96 which makes sense because that is the
velocity function.
We took the derivative.
We took the velocity function.
When we took the integral of the velocity function, we got the position function.
So we just went backwards again.
We take the derivative.
We set that equal to zero and now we solve for t by adding -32t to both sides and then
dividing both sides by 32 and we see that t is equal to 3.
So knowing the t = 3, we can now solve for y. and the way that we’re going to do that
is by plugging in 3 into the position function.
So y(3) = -16(3^2) + 96(3).
And when we solve that y(3), we get -144 + 288.
When we add those together, we get a positive 144 feet.
So this is the maximum height that the ball attains.
So that was the first question: How high does the ball rise?
Now we have to find the time t when it hits the ground because that’s what will tell
us how long it remains aloft.
We only know here that the height of the ball is equal to zero.
so if we set the position function equal to zero, we get 0 = -16t^2 + 96t.
we can factor this and then solve for t.
So we’re going to pull out a negative 16t and then we’ll multiply by t – 6.
So solving this, we can see that t is either going to be equal to zero or 6.
We already know that the height of the ball is zero at 0.
We know that the height is zero when t is equal to zero.
But we’ve also found that the height is zero when t is equal to 6.
Which means that t = 6 and that’s 6 seconds.
So those are our two answers to this problem.
The ball reaches a maximum height of 144 feet.
We know that it occurs at time t = 3.
And in terms of how long it remains aloft, we know that it stays in the air for 6 seconds
before it hits the ground.
So that occurs at time t = 6.
So that’s it.
I hope this video helped you guys and I will see you in the next one.
Bye!