Question 1159348
given the function: 

{{{h=40t-5t^2}}} 


1).what is the height of the arrow in meters after {{{5}}} seconds?

{{{h=40t-5t^2}}} ........if {{{t=5}}} seconds

{{{h=40*5-5*5^2}}}
 
{{{h=200-125}}} 

{{{h=75}}}  meters high


2). After how many seconds will the arrow be {{{30}}} meters high? 

{{{h=40t-5t^2}}} ........if {{{h=30}}} 

{{{30=40t-5t^2}}}

{{{5t^2-40t+30=0}}}......simplify, divide by {{{5}}}

{{{t^2-8t+6=0}}}.......use quadratic formula

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

{{{t = (-(-8) +- sqrt( (-8)^2-4*1*6 ))/(2*1) }}} 

{{{t = (8 +- sqrt(64-24))/2 }}} 

{{{t = (8 +- sqrt(40))/2 }}} 

{{{t = (8 +- sqrt(4*10))/2 }}}

 {{{t = (8 +- 2sqrt(10))/2 }}} .........simplify

 {{{t = (4 +- sqrt(10)) }}}

-> solutions:

{{{t = (4 + sqrt(10)) }}}->{{{t = 7.16 }}} seconds

{{{t = (4 - sqrt(10)) }}}->{{{t = 0.84}}} seconds


so, it will ritch height of {{{30}}} meters in {{{ 0.84}}} seconds and second time (on its way back) {{{t = 7.16 }}} seconds


{{{drawing ( 600, 600, -10, 45, -10, 45,
circle(0.84,30,.13),circle(7.16,30,.13),locate(2,31,h=30m),

graph( 600, 600, -10, 45, -10, 45, 40x-5x^2,30)) }}}



3). when will the arrow be back on the ground? 

the arrow be back on the ground when {{{h=0}}}

{{{0=40t-5t^2}}}

{{{0=5t(8-t)}}}

->{{{5t=0}}}->when {{{t=0}}}...this is start position

->{{{8-t=0}}}->when {{{t=8}}}-> it will take {{{8}}} seconds for the arrow to be back on the ground