Question 1159584


 
The height {{{h }}}of a missile after {{{t}}} seconds, when fired straight up with an initial height of {{{2}}} feet, can be modeled by the function 

{{{h(t)=-16t^2+150t+2 }}}

A.) When will the missile be {{{250 }}}feet?

->{{{h=250 }}}

{{{250=-16t^2+150t+2 }}}

{{{16t^2-150t-2 +250=0}}}

{{{16t^2-150t+248=0}}}......use quadratic formula


{{{t = (-b +- sqrt( b^2-4*a*c ))/(2*a) }}} 

{{{t = (-(-150) +- sqrt( (-150)^2-4*16*248 ))/(2*16) }}} 

{{{t = (150 +- sqrt( 22500-15872 ))/32 }}} 

{{{t = (150 +- sqrt( 6628 ))/32 }}} 

{{{t = (150 +- 81.413)/32 }}} 

solutions:

{{{t = (150 + 81.413)/32 }}} ->{{{t =7.23 }}}

{{{t = (150 - 81.413)/32 }}} ->{{{t =2.14 }}}

fired straight up the missile will be {{{250 }}} feet high in {{{t =2.14 }}} seconds, and on way back again in {{{t =7.23 }}} seconds



B.) When will the missile be at it's maximum height?

 if {{{a}}} is negative, the parabola opens down and maximum is at vertex

so, write your equation in vertex form:

{{{h(t)=-16t^2+150t+2 }}}

{{{h(t)=-16(t^2-(150/16)t)+2 }}}.....complete square

{{{h(t)=-16(t^2-(75/8)t+b^2)-(-16)b^2+2 }}}......{{{b=(75/8)/2=(75/16)}}}

{{{h(t)=-16(t^2-(75/8)t+(75/16)^2)+16(75/16)^2+2 }}}

{{{h(t)=-16(t-75/16)^2+5657/16}}}

{{{h(t)=-16(t-75/16)^2+353.5625}}}

-> {{{h=75/16=4.7}}} and {{{k=353.6}}}

vertex: ({{{4.7}}},{{{353.6}}})

the missile will be at it's maximum height in {{{4.7}}} seconds



C.) What is the maximum height of the missile?

 {{{h=353.6ft}}}


D.)Will the missile ever reach a height of {{{400ft}}}?

{{{h=400ft}}}

{{{400=-16t^2+150t+2 }}}

{{{16t^2-150t-2 +400=0}}}

{{{16t^2-150t +398=0}}}.......check discriminant

If {{{b^2−4ac < 0}}} the equation has {{{no}}} real number solutions, but it does have complex solutions

{{{(-150)^2-4*16*39<0}}}
{{{-2972< 0}}}-> true

so, the missile will never reach a height of {{{400ft}}}