Question

The height in feet of a rocket after t seconds is given by h(t) = 160t - 16t^2.

Find the maximum height the rocket attains in feet​

Answer 1

Answer:

Step-by-step explanation:

h(t)=160t-16t^2=-16(t²-10t+25-25)

=-16(t-5)²+400

max. height reached=400 ft

Answer 2

Answer:

The maximum height 400 feet is attained at t = 5 seconds.

Step-by-step explanation:

All powers of $t$ in the equation for $h(t)$ are integers that are greater than or equal to zero. Additionally, the greatest power of $t$ is two. Hence, $h(t)$ is a quadratic equation about $t$.

• Let $a$ be the coefficient of the $t^2$ term, and
• Let $b$ be the coefficient of the $t$ term.

In this case,

• $a = -16$, and
• $b = 160$.

The question is asking for the maximum value of this equation. Start by finding the $t$ (time) that would maximize the value of the polynomial.

The graph of a quadratic equation looks like a parabola. Additionally, since the coefficient of $t^2$ is less than zero, the parabola opens downward. The maximum value of the parabola would be at its vertex. Additionally, at the vertex,

$\displaystyle t = -\frac{b}{2a} = -\frac{160}{2 \times (-16)} = \frac{160}{32} = 5$.

In other words, the rocket is at its maximum height when time is equal to 5 seconds.

To find that height, let $t = 5$ and evaluate $h(t) = 160 \, t - 16\, t^2$:

$160\, t - 16\, t^2 = 160 \times 5 - 16 \times 5^2 = 400$.

That is: the maximum height of the rocket would be 400 feet.