Question

A path of a toy rocket thrown upward from the ground at a rate of 208 ft/sec is modeled by the quadratic function h(t) = -16t +208t. When will the rocket reach its maximum height? What will be the maximum height?

Answer 1

Answer:

The time = 6.5 seconds and the maximum height reached by the rocket = 676 feet

Step-by-step explanation:

A path of a toy rocket thrown upward from the ground at a rate of 208 ft/sec is modeled by the quadratic function h(t) = -16$t^{2}$ +208t.

We have to find when the rocket will reach its maximum height.

When the rocket reaches its maximum height, $\frac{dh}{dt}$ = 0.

h(t) = -16$t^{2}$ +208t

$\frac{dh}{dt}$ = - 32t + 208 = 0

t = 6.5 seconds

At t = 6.5 seconds it is at maximum height.

Maximum height = - 16$\times (6.5^{2} ) + 208\times 6.5$

                            = 676 feet

Answer 2

Answer:

Time taken = 6.5s . Maximum height = 676 feet.

Step-by-step explanation:

To find the maxima of height function use differentiation ie, h(t) will be maximum when $\frac{dh(t)}{dt} = 0$ .

$h(t) = -16t^{2} + 208t\\\frac{dh(t)}{dt}= -32t + 208$

$\frac{dh(t)}{dt} = 0$ ⇒ -32t +208 = 0

32t = 208

$t = \frac{208}{32} = 6.5$

For maximum height put t = 6.5  in h(t),

Maximum height = -16×6.5×6.5  + 208×6.5

=676 feet.