Question 1156441

The height of a projectile fired upward from the ground with a velocity of 128 feet per second is given by the formula: 

{{{h=-16t^2 + 128t}}}

a. When will the projectile be {{{256}}} feet above the ground?

{{{h=256}}}
{{{256=-16t^2 + 128t}}}
{{{16t^2-128t+256=0 }}}....both sides divide by {{{16}}}
{{{t^2-8t+16=0 }}}......factor
{{{t^2-4t-4t+16=0 }}}
{{{(t^2-4t)-(4t-16)=0 }}}
{{{t(t-4)-4(t-4)=0 }}}
{{{(t-4)(t-4)=0 }}}

=>{{{t=4}}}
the projectile will be {{{256}}} feet above the ground in {{{4}}}  seconds



b. Will the projectile ever be {{{300}}} feet above the ground? Explain.
{{{h=300}}}
{{{300=-16t^2 + 128t}}}
{{{16t^2-128t+300=0 }}}

use quadratic formula:

{{{t=(-b+-sqrt(b^2-4ac))/2a}}}...in your case {{{a=16}}},{{{b=-128}}}, and {{{c=300}}}
{{{t=(-(-128)+-sqrt((-128)^2-4*16*300))/(2*16)}}}
{{{t=(128+-sqrt(16384-19200))/32}}}
{{{t=(128+-sqrt(-2816))/32}}}=> since determinant is negative number, we have imaginary roots 
=> the projectile will never be {{{300}}} feet above the ground


c. When will the projectile be {{{240}}} feet above the ground?

{{{h=240}}}
{{{240=-16t^2 + 128t}}}
{{{16t^2-128t+240=0 }}}
{{{16(t^2-8t+15)=0 }}}

use quadratic formula:
{{{t=(-b+-sqrt(b^2-4ac))/2a}}}...in your case {{{a=1}}},{{{b=-8}}}, and {{{c=15}}}
{{{t=(-(-8)+-sqrt((-8)^2-4*1*15))/(2*1)}}}
{{{t=(8+-sqrt(64-60))/2}}}
{{{t=(8+-sqrt(4))/2}}}

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

solutions:
{{{t=(8+2)/2}}}=>{{{t=5}}}
{{{t=(8-2)/2}}}=>{{{t=3}}}

 the projectile will be {{{240}}} feet above the ground in {{{3}}} and again in {{{5}}}

d. In how many seconds will the projectile hit the ground?

{{{h=0}}}
{{{0=-16t^2 + 128t}}}

{{{16t^2 -128t=0}}}

{{{16t(t^2 -8)=0}}}

if {{{16t=0}}}=>{{{t=0}}}=> not solution

if  {{{t^2 -8=0}}}=> {{{t^2 =8}}}=> {{{t =sqrt(8)}}}=> {{{t =2.83}}}

the projectile hit the ground in {{{2.83}}} seconds