Question 1129339


{{{E(t)}}} is just a parabola. I'm sure you know how to find the vertex, where {{{E(t)}}} is a maximum.

{{{E(t) = -279.67t^2 + 82.86t}}}
where 0 ≤ t ≤ 0.3

{{{E(t) = -279.67t^2 + 82.86t}}}......complete square to get vertex form

{{{E(t) = -279.67(t^2 +( 82.86/-279.67)t+b^2)-279.67(-b^2)}}}

{{{E(t) = -279.67(t^2 -0.2962778t+b^2)-279.67(-b^2)}}}

{{{b=0.2962778/2=0.148139}}}

{{{E(t) = -279.67(t^2 -0.2962778t+0.148139^2)-279.67(-0.148139^2)}}}
{{{E(t) = -279.67(t -0.148139)^2+6.1374}}}

max ate vertex: ({{{t}}}, {{{E(t)}}})=({{{0.148139}}}, {{{6.1374}}})

so, {{{E(t)=6.1374}}}, round to the nearest tenth of a joule:

{{{6.1}}} joule