Respuesta :
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[tex]t^{2}[/tex] +208t.
We have to find when the rocket will reach its maximum height.
When the rocket reaches its maximum height, [tex]\frac{dh}{dt}[/tex] = 0.
h(t) = -16[tex]t^{2}[/tex] +208t
[tex]\frac{dh}{dt}[/tex] = - 32t + 208 = 0
t = 6.5 seconds
At t = 6.5 seconds it is at maximum height.
Maximum height = - 16[tex]\times (6.5^{2} ) + 208\times 6.5[/tex]
= 676 feet
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 [tex]\frac{dh(t)}{dt} = 0[/tex] .
[tex]h(t) = -16t^{2} + 208t\\\frac{dh(t)}{dt}= -32t + 208[/tex]
[tex]\frac{dh(t)}{dt} = 0[/tex] ⇒ -32t +208 = 0
32t = 208
[tex]t = \frac{208}{32} = 6.5[/tex]
For maximum height put t = 6.5 in h(t),
Maximum height = -16×6.5×6.5 + 208×6.5
=676 feet.
