Question 1169625
a) How long is the ball in the air for?

{{{H= -5t^2+20t}}}-> t is time in seconds, so set H=0 and solve for t
{{{0= -5t(t-4)}}}
if {{{0= -5t }}}->{{{t=0}}} 
or
if {{{0= t-4}}}->{{{t=4}}}-> your solution is: the ball is in the air for {{{t=4}}} seconds

b) When will the ball reach its maximum height?
trajectory is a parabola and maximum is vertex
write {{{H= -5t^2+20t}}} in vertex form, complete square
{{{H= -5t^2+20t}}}...first factor out {{{-5}}}
{{{H= -5(t^2-4t)}}}... complete square
{{{H= -5(t^2-4t+b^2)-(-5b^2)}}}...{{{b=4/2=2}}}
{{{H= -5(t^2-5t+2^2)+5(2)^2}}}
{{{H= -5(t-2)^2+5(4)}}} 
{{{H= -5(t-2)^2+20}}}

vertex is at ({{{20}}}, {{{2}}})=> the ball reach its maximum in {{{20}}} seconds

c) What is the maximum height reached by the ball?
the  maximum height reached by the ball is {{{2m}}} 

d) At what time(s) will the ball be {{{15}}} meters in the air?

{{{H= 15m}}}

{{{15= -5t^2+20t}}} ....solve for {{{t}}}

{{{15+5t^2-20t=0}}} 

{{{5t^2-20t+15=0}}} 

{{{5(t^2-4t+3)=0}}}...will be zero if

 {{{t^2-4t+3=0}}}....factor

{{{t^2-3t-t+3=0}}}

{{{(t^2-3t)-(t-3)=0}}}

{{{t(t-3)-(t-3)=0}}}

{{{(t - 3) (t - 1) = 0}}}

=> {{{t=3}}} , {{{t=1}}}



{{{ drawing( 600, 600, -10, 10, -10, 25,
circle(3,15,.12),circle(1,15,.12),locate(3,15,p(3,15)),locate(1,15,p(1,15)),
 graph( 600, 600, -10, 10, -10, 25, -5x^2+20x, 15) )}}}