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