Question

A rocket is 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=−16t^2+96t+2560.
What is the rocket's maximum height?

How long does it take for the rocket to reach its max height?

After it's fired, the rocket reaches the ground at t=__ sec.

Answer 1

Check the picture below.

$\bf \textit{ vertex of a vertical parabola, using coefficients}\\\\ \begin{array}{lccclll} h(t) = &{{ -16}}t^2&{{ +96}}t&{{ +2560}}\\ &\uparrow &\uparrow &\uparrow \\ &a&b&c \end{array}\qquad \left(-\cfrac{{{ b}}}{2{{ a}}}\quad ,\quad {{ c}}-\cfrac{{{ b}}^2}{4{{ a}}}\right)$

so.. the maximum point is at   $\bf {{ c}}-\cfrac{{{ b}}^2}{4{{ a}}}~~~ft$

and it happened after   $\bf -\cfrac{{{ b}}}{2{{ a}}}~~~seconds$

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now, when the rocket reaches the ground, as you saw on the picture, y = 0, thus

$\bf h(t)=-16t^2+96t+2560\implies 0=-16t^2+96t+2560 \\\\\\ 0=-16(t^2-6t-160)\implies 0=t^2-6t-160 \\\\\\ 0=(t-16)(t+10)\implies t= \begin{cases} \boxed{16}\\ -10 \end{cases}$

"t" is an amount in seconds, for this specific phenomena, it cannot be negative, so -10 is not a feasible value.

Answer 2

Answer:

Rocket's maximum height = 2704 ft

It takes 3 seconds to reach maximum height.

The rocket reaches the ground at 16 seconds.

Step-by-step explanation:

The equation of height is given by h = −16t²+96t+2560

At maximum height the derivative of the above equation is zero.

That is

         -32t + 96 = 0

              t = 3 s

So, It takes 3 seconds to reach maximum height.

At t = 3s

       h = −16 x 3²+96 x 3+2560 = 2704 ft

Rocket's maximum height = 2704 ft

When the rocket reaches ground its height is zero

          −16t²+96t+2560 = 0

             t²-6t-160 = 0

             (t-16)(t+10) = 0

              t = 16 s or t = -10 s (not possible)

The rocket reaches the ground at 16 seconds.