Question

A rocket is fired directly upwards with a velocity of 96 ft/sec, from a platform which is 11 feet above the ground. The equation for the rocket's height, H , as a function of time, t , is given by the function
H(t)=-16t^2+96t+11

Answer 1

ok

Answer 2

I have to find the maximum height

Answer 3

Do you know what the vertex of a quadratic?

Answer 4

is*

Answer 5

the max or min of the function

Answer 6

do you know how to find it?

Answer 7

not algebraically

Answer 8

for the quadratic \(ax^2+bx+c\) the \(x\) coordinate of the vertex is \(x=\dfrac{-b}{2a}\). To find the \(y\) value, which is what you are asked for, plug that into the polynomial.

Answer 9

H(t)=-16t^2+96t+11
so H(t)= 16(-b/2a)^2+96(-b/2a)+11?

Answer 10

-96/32=-3

Answer 11

\(\dfrac{-b}{2a}=\dfrac{-96}{-32}=3\)

Answer 12

now plug that into the quadratic to get the y value

Answer 13

thanks

Answer 14

np