Question 120868
{{{s(t)=-16t^2+v[0]t+s[0]}}}


a.  You are given that {{{v[0]=32}}} and {{{s[0]=40}}}, so just plug in the values:


{{{s(t)=-16t^2+32t+40}}}


b.  You need to evaluate {{{s(0.5)}}}


{{{s(0.5)=-16(0.5)^2+32(0.5)+40}}}
{{{s(0.5)=-16(0.25)+32(0.5)+40}}}
{{{s(0.5)=-4+16+40}}}
{{{s(0.5)=52}}}


So the height (measured from the ground) after one-half second is 52 feet.


c.  This is a quadratic equation and if you graphed it on a coordinate plane with s as the vertical axis and t as the horizontal axis, you would have a convex down parabola.  The maximum height will be reached at time equal to the value of t at the vertex of the parabola.


The vertex of any parabola described by {{{f(x)=ax^2+bx+c}}} is located at <b><big>({{{(-b)/2a)}}},{{{f((-b)/2a)}}})</b></big>


For this problem, {{{a=-16}}} and {{{b=32}}}, hence {{{t[maxh]=(-32)/(2*(-16))=(-32)/-32=1}}} second.


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You can also do this part with the calculus


A local minimum or maximum is found where the first derivative equals 0.


For this problem, s'(t)={{{-32t+32}}}, so if {{{-32t+32=0}}} then {{{x=1}}}


To determine if this is a maximum or minimum, evaluate the second derivative at the same value


s"(t)={{{-32}}}.  Since the second derivitive is less than zero, this is a maximum.


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d. The actual maximum height is just the function evaluated at the time for maximum height, i.e. 1 second, or {{{s(1)}}}


{{{s(1)=-16(1)^2+32(1)+40}}}
{{{s(1)=-16+32+40}}}
{{{s(1)=56}}}


And the maximum height is 56 feet.