Question

model rocket is launched from a raised platform at a speed of 176 feet per second. Its height in feet is given by h(t)=-16t^2+176t+20 (t=seconds after launch)
What is the maximum height reached by the rocket? and what time does it meet maximum height?

Answer 1

The first step is to find the vertex. You do this by using the formula $\frac{-b}{2a}$. 

Plug in b and a into the equation: $\frac{-176}{2(-16)}$

You get $5.5$

So, it takes 5.5 seconds to get to 504 feet

Answer 2

Hi, apply the formula:

Xv = - b / 2a

Where,

b = 176
a = -16
c = 20

Then, the value of Xv will be:

Xv = - (176 / 2 .-16)

Xv = 176 / 32

Xv = 5,5 s

So , just us will go to make the substitute of Xv in equation.
And we will go to find the value of H

H(t) = Yv

As , Yv = h(Xv)

H(t) = -16t^2 + 176t +20

Yv = - 16.(5,5)^2 + 176.(5,5)+20

Yv = -484 + 968 + 20

Yv = 504 feet

This would be the answer to the your question.

Hope this helps