Question 1025292
{{{H(t) =-16t^2 + 240t + 4000}}} defined for the time that the arrow is in flight,
{{{t>=0}}}  and {{{t}}} such that {{{H(t)>=0}}} ,
is the height of the arrow above the canyon floor in feet {{{t}}} seconds after being shot.
 
(a) {{{H(3)=-16*3^2 + 240*3 + 4000=-16*9+720+4000=-144+720+4000=4578}}} 
After {{{3}}} seconds, the arrow was at a height of {{{4578}}} feet. 
 
(b){{{H(t) =1536}}}
{{{-16t^2 + 240t + 4000=1536}}}
{{{-16t^2 + 240t + 4000-1536=0}}}
{{{-16t^2 + 240t + 2464=0}}} .
Dividing both sides of the equal sign by {{{-16}}} we get the equivalent equation
{{{t^2-15t-154=0}}} , which we can solve by factoring:
{{{t^2-15t-154=0}}}-->{{{(t+7)(t-22)=0}}}-->{{{system(t+7=0,"or",t-22=0)}}}-->{{{system(t=-7,"or",t=22)}}} .
{{{t=-7}}} does not make sense; it is not in the domain of the function, where it must be {{{t>=0}}} ,
so the answer is {{{t=22}}} .
It took about {{{22}}} seconds for the arrow to get to 1536 feet.
 
c) When the arrow hits the canyon floor, it is {{{0}}} feet above the canyon floor, so
{{{H(t)=0}}} , which means {{{-16t^2 + 240t + 4000=0}}}
Dividing both sides of the equal sign by {{{-16}}} we get the equivalent equation
{{{t^2-15t-250=0}}} , which we can solve by factoring:
{{{t^2-15t-250=0}}}-->{{{(t+10)(t-25)=0}}}-->{{{system(t+10=0,"or",t-25=0)}}}-->{{{system(t=-10,"or",t=25)}}} .
{{{t=-10}}} does not make sense; it is not in the domain of the function, where it must be {{{t>=0}}} ,
so the answer is {{{t=25}}} .
After {{{25}}} seconds, the arrow hits the canyon floor.