Question 1153455


NASA launches a rocket at {{{t=0}}} seconds. Its height, in meters above sea-level, as a function of time is given by 

{{{h(t)=-4.9t^2+217t+185}}}

Assuming that the rocket will splash down into the ocean, at what time does splashdown occur?

will occur when {{{h(t)=0}}}

{{{0=-4.9t^2+217t+185}}}.........use quadratic formula

{{{t = (-217 +- sqrt( 217^2-4*(-4.9)*185 ))/(2*(-4.9)) }}}

{{{t = (-217 +- sqrt( 47089+3626 ))/(-9.8) }}}

{{{t = (-217 +- sqrt( 50715))/(-9.8) }}}

{{{t = (-217 +- 225.1999)/(-9.8) }}}

solutions: we need only positive solution

{{{t = (-217 - 225.1999)/(-9.8) }}}

{{{t = -(217+225.1999)/(-9.8) }}}

{{{t = (217+225.1999)/9.8 }}}

{{{t =45.1224}}}



The rocket splashes down after ___{{{t =45.1224}}}_____ seconds.


How high above sea-level does the rocket get at its peak?

{{{h(t)=-4.9t^2+217t+185}}}

max occurs when {{{t = -b/(2a)=-217/(2(-4.9))=-217/-9.8=22.14 }}}

{{{h(22.14 )=-4.9(22.14 )^2+217*22.14 +185}}}

{{{h(22.14 )=2587.5}}}

The rocket peaks at ____{{{2587.5}}}______ meters above sea-level.