Thread: Max Height of Ball

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

Hi,

Originally Posted by magentarita

the question mentions that the ball is thrown upward at 48 feet per second.

Yes but as the ball is thrown upward its speed decreases by 32 ft/s each second because of the gravitational acceleration (g=32 ft/s^2). Your answer does make sense.

Originally Posted by flyingsquirrel

Hi,

Yes but as the ball is thrown upward its speed decreases by 32 ft/s each second because of the gravitational acceleration (g=32 ft/s^2). Your answer does make sense.

Can you further explain your reply? I placed this question in another math forum and the person who replied said my answer is right.

Originally Posted by magentarita

the person who replied said my answer is right.

That's also what I said :

Originally Posted by flyingsquirrel

Your answer does make sense.

Originally Posted by flyingsquirrel

That's also what I said :

I got it now. Thanks.