Question

A rocket fired upward from some initial distance above the ground. Its height (in feet) h, above the ground t seconds after it is fired is given by h(t)=-16t^2+160t+384

What is the rockets maximum height?


After it is fired, The rocket reaches the ground at t=____ seconds

Answer 1

Answer:

Part A)

784 feet in the air (after five seconds).

Part B)

After 12 seconds.

Step-by-step explanation:

The height h (in feet) of a rocket t seconds after being fired is modeled by the function:

$h(t)=-16t^2+160t+384$

Part A)

We want to find the rocket's maximum height.

Since our function is a quadratic, the maximum height will occur at its vertex. The vertex of a quadratic is given by:

$\displaystyle \Big(-\frac{b}{2a}\, f\Big(-\frac{b}{2a}\Big)\Big)$

In this case, a = -16, b = 160, and c = 384.

So, the vertex occurs at:

$\displaystyle t=-\frac{160}{2(-16)}=\frac{160}{32}=5\text{ seconds}$

The maximum height is reached after five seconds.

Then the maximum height is:

$h(t)_\text{max}=h(5)=-16(5)^2+160(5)+384=784\text{ feet}$

Part B)

When the rocket reaches the ground, its height h above the ground will be 0. Hence:

$h(t)_\text{ground}=0=-16t^2+160t+384$

Solve for t. We can first divide both sides by -16:

$t^2-10t-24=0$

Factor:

$(t-12)(t+2)=0$

Zero Product Property:

$t-12=0\text{ or } t+2=0$

Solve for each case:

$t=12\text{ or } t=-2$

Time cannot be negative. Hence, our only solution is:

$t=12\text{ seconds}$

The rocket reaches the ground 12 seconds after it is fired.