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

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

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,
`(dh)/(dt)` = 0.

h(t) = -16
`t^(2)` +208t

`(dh)/(dt)` = - 32t + 208 = 0

t = 6.5 seconds

At t = 6.5 seconds it is at maximum height.

Maximum height = - 16
`* (6.5^(2) ) + 208* 6.5`

= 676 feet

Answer 2

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

Explanation:

To find the maxima of height function use differentiation ie, h(t) will be maximum when
`(dh(t))/(dt) = 0` .

`h(t) = -16t^(2) + 208t\\(dh(t))/(dt)= -32t + 208`

`(dh(t))/(dt) = 0` ⇒ -32t +208 = 0

32t = 208

`t = (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.