Question

Please help with this question.
A rocket leaves a launcher at a height of 6 feet off the ground with an initial velocity of 128 feet per second. The equation describing the rocket's height after t seconds is h(t)=-16t^2+128t+6. Find the maximum height reached by the rocket and how many seconds it takes for the rocket to reach that height

Answer 1

Given:

The equation describing the rocket's height after t seconds is

$h(t)=-16t^2+128t+6$

To find:

The maximum height and time when it is maximum.

Solution:

We have,

$h(t)=-16t^2+128t+6$

Leading coefficient is -16, which is negative. So, it is a downward parabola.

Vertex of a downward parabola is the point of maxima.

If a quadratic function is $f(x)=ax^2+bx+c$, then

$Vertex =\left(\dfrac{-b}{2a},f\left(\dfrac{-b}{2a}\right)\right)$

In the given function, a=-16, b=128 and c=6.

$\dfrac{-b}{2a}=\dfrac{-128}{2(-16)}$

$\dfrac{-b}{2a}=4$

Putting x=4 in the given function, we get

$h(4)=-16(4)^2+128(4)+6$

$h(4)=-16(16)+512+6$

$h(4)=-256+518$

$h(4)=262$

So, vertex of the function is (4,262).

Therefore, maximum reached by the rocket is 262 feet it 4 seconds.