Question 1160350

 {{{H=-16t^2+336t }}}where {{{t}}} is the time in seconds.
When does the ash projectile reach its maximum height?

It's the vertex of the parabola at {{{t = -b/2a}}}
in your case {{{b=336}}} and {{{a=-16}}}

then

 {{{t = -336/(2(-16))}}}

 {{{t = -336/(-32)}}}

 {{{t = 336/32}}}

{{{t =21/2}}}



What is its maximum height?
Sub {{{t }}}at the vertex.

{{{H=-16(21/2)^2+336(21/2) }}}

{{{H=-16(441/4)+3528}}}

{{{H=-1764+3528}}}

{{{H=1764}}}

max {{{H= 1764 }}}at {{{t = 21/2}}}


when does the ash projectile return to the ground?
When {{{-16t^2 + 336t = 0}}}

Which is {{{2}}} times the{{{ t}}} at the vertex
{{{t }}}at the vertex: {{{t =21/2}}}
 {{{2}}} times the {{{t}}} at the vertex: {{{t=2(21/2)=21}}}

the ash projectile return to the ground in {{{21}}} seconds