Question 1088660


{{{s(t) = -16t^2 + 800t + 600}}} 

After how many seconds will the projectile be {{{5000ft}}} above the ground=>{{{s(t) =5000ft}}}

{{{5000= -16t^2 + 800t + 600}}} 

{{{5000+16t^2 - 800t -600=0}}} 

{{{16t^2 - 800t+4400=0}}} 

{{{t^2 - 50t+275=0}}} 

{{{t^2 - 50t+275=0}}} 


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

{{{t = (-(-50) +- sqrt( (-50)^2-4*1*275 ))/(2*1) }}}

{{{t = (50 +- sqrt( 2500-1100 ))/2 }}}

{{{t = (50 +- sqrt(1400 ))/2 }}}

{{{t = (50 +- sqrt(14*100 ))/2 }}}

{{{t = (50 +- 10sqrt(14 ))/2 }}}

{{{t = (25 +- 5sqrt(14 )) }}}

{{{t = 5(5 +- sqrt(14 )) }}}

solution:

{{{t = 5(5 + sqrt(14 )) }}}

{{{t = 5(5 + 3.741657386773941) }}}

{{{t = 5(8.741657386773941) }}}

{{{t = 43.7 }}}

the projectile be {{{5000}}} ft above the ground after approximately {{{ 43.7 }}} seconds