Question 1139880
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It will be at the given height when  h(t) = 380,  or, equivalently,


    {{{-16t^2 + 160t}}} = 380


    {{{16Tt^2 -160t + 380}}} = 0.


Solve this quadratic equation


    {{{t[1,2]}}} = {{{(160 +- sqrt(160^2 - 4*16*380))/(16*2)}}} = {{{(160 +- 35.78)/32}}}.


There are two solution, both make sense, {{{t[1]}}} = {{{(160 - 35.78)/32}}} = 3.88 seconds in the increasing part of the parabola,

and  {{{t[2]}}} = {{{(160 + 35.78)/32}}} = 6.12 seconds in the decreasing branch of the parabola.
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Solved.