# How do I work out this math equation?

Hello :)

A ball is thrown upward from the ground level. Its height h, in feet, above the ground after t seconds is h=48t-16t^2. Find the maximum height of the ball.

I already know that the answer is 36 (max height). But I’m not sure how to get there.. can you help explain it to me? Thank you!

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## 14 Answers

Is this for a calculus class? Unless you apply some physics concepts, you need to use calculus to solve this.

You need to optimize h. Start by taking the derivative of the equation.

The easiest way is using calculus. To find a maximum or minimum you look for a point where the tangent to the curve is horizontal, which implies that the derivative equals zero (“vanishes” in the parlance).

If you don’t know, or aren’t supposed to know, how to differentiate a function then this might be a bit more complicated. Do you know about parabolas?

Once you find a value of t when the height is maximum, you plug t back into the original equation of motion to find corresponding height h.

@gasman

But how do I find the value of t in this equation? Once I get that part figured out I’ll be ok.

@kittykat219 Are you expected to know how to do derivatives? Take the derivative of the equation and set it equal to zero, then solve for t.

Yeah I think I’m supposed to know how to do them haha.

So the equation would then turn to 48t-16t^2=0 right?

But how can you then simplify that down so that you only have t= ?

Wait, the derivative is 48–32t?

How do you figure out derivatives?

There is a **much, much easier way** to do this than calculus.

First, the path of the ball follows a parabola. To find the maximum point (the vertex) of the parabola, the x-coordinate of the vertex is the point directly between the two x-intercepts of the parabola. You then plug this number to find the y-coordinate.

First, you take the equation 48t-18t^2 and set it to zero.

0=48t-16t^2

0=t(48–16t)

For this to be equal to 0, either t is zero or 48–16t=0.

48=16t

t=3

The x-ints are 0 and 3. This means that the x coordinate is 1.5 for the vertex, plug this in to the formula and you get y=36.

Do you have a graphing calculator for your class? You can graph the function and use the calculator to find the maximum. Your teacher may or may not find that acceptable.

Or what @PhiNotPi just did.

Yes, you’ve got the right derivative in your last reply.

For polynomial functions, for each term, multiple the coefficient by the exponent and subtract one from the exponent to get the derivative.

What @PhiNotPi gave is also right, but if this is a calculus class I don’t think your teacher will accept a non-calculus solution.

@PhiNotPi

Ok, so you get t=3, but how do you then come up with 1.5?

I don’t think my teacher will mind me doing it this way, it’s for algebra class :)

Oh! That changes everything.

1.5 is directly in between 0 and 3. 0 and 3 are the x intercepts, so the vertex (the highest point) will always be exactly in between (for a regular parabola, which this is).

This may help you visualize it. Click on “properties” in the geometric figures section to see why 36 is the maximum.

Always good to visualize a graph, as supplied by @cockswain. I agree with @PhiNotPi that this problem is easily solved using the symmetry of a parabola— & if this is algebra class it’s probably what you’re supposed to do.

Calculus, as you will perhaps learn, provides a simple, general method for this kind of problem. Here you would solve 48 – 32t = 0.

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