The height in feet for a ball thrown upward at 48 feet per second is given by

s(t)= - 16t^2 + 48t. Where t is the time in seconds after the ball is tossed. What is the maximum height that the ball will reach?

Here is my work:

This is a parabola. It's a quadratic (the highest exponent is 2). The -16 tells me that this parabola opens downward.

The maximum height of the ball will be found at the vertex of the parabola.

I have this function:

s(t) = -16t^2 + 48t.

I first find the vertex of a parabola using the following formula:

t = -b/2a, where b = -48 and a = -16.

I plug this into the formula t = -b/2a and simplify.

t = -48/2(-16)

t = -48/-32

t = 48/32

t = 3/2

I used the formula above to find t.

Now, to find the height of this ball, I decided to plug 3/2 into the given function and simplify.

s(t) = -16^2 + 48t

I let t = 3/2

s(3/2) = -16(3/2)^2 + 48(3/2)

s(3/2) = -36 + 72

s(3/2) = 36

The maximum height is 36 feet.

Of course, this does not make sense because the question mentions that the ball is thrown upward at 48 feet per second.

What did I do wrong?

Thanks