Question 1137680

Gloria is skeet shooting. The height of the skeet is modelled by the function 

{{{h(t)=-4.9t^2+32t+2}}} 

where {{{h(t)}}} is the height in meters after {{{t}}} seconds.

 
The path of Gloria’s pellet is modeled by the function 

{{{g(t)=28.5t+1}}} , with the same units. 



a) How high off the ground will the skeet be when it is hit?
 
when {{{h(t) =g(t) }}}, and it will be at the point where {{{h(t) }}}and{{{ g(t) }}}}intersects 


first find the time when
  {{{h(t) =g(t) }}}

{{{-4.9t^2+32t+2=28.5t+1}}}

{{{0=4.9t^2+28.5t-32t+1-2}}}

{{{4.9t^2-3.5t-1=0}}}

{{{t}}}≈{{{-0.218733}}} seconds->Reject the negative solution

{{{t}}}≈{{{0.933018}}} seconds


now find height

{{{h(t)=-4.9(0.933018)^2+32(0.933018)+2}}}

{{{h(t)=-4.2655606827876+29.856576+2}}}

{{{h(t)=27.59}}}


the point where {{{h(t)}}} and {{{g(t) }}}intersects  is 

({{{0.933018}}}, {{{27.59}}})


b) After how many seconds will the skeet be hit? 

after {{{t=0.933018}}} seconds

 it will take {{{0.933018 }}}seconds to hit the skeet and the skeet will be
{{{27.59}}} meters in the air