Question

The height s of a ball, in feet, thrown with an initial velocity of 112 feet per second from an initial height of 20 feet is given as a function of time t (in seconds) by
s(t)= -16t^2+112t+20. What is the maximum height of the ball? At what time does the maximum height occur?

I know how to find the maximum height. You find the vertex which is 3.5 then plug it back into the original equation.

Now, when it says what time, I am not sure if it wants the vertex or should I use the quadratic to solve it?

Please help.

Answer 1

The maximum height <i>is</i> the vertex.

This would be an application of completing the squares to convert the function into the standard form:
s(t)=a(t-h)²+k
where t=h at the vertex (3.5) and
k=maximum height (216 above datum)

Answer 2

Ok thanks.

Answer 3

You're welcome!