Question 176994
{{{s(t) = -16t^2 + v[o]t + s[o]}}}


The initial velocity {{{v[o] = 32}}} is a given ("...thrown upward with the initial velocity of 32 feet per secound[sic]...")


The initial height {{{s[o] = 40}}} is a given ("...from the top of a 40 foot building...")


So just just substitute:


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


Now, the question is what is the height at 5 seconds.  {{{s(t)}}} is the height function with respect to time, so just substitute 5 for t and do the arithmetic.


{{{s(5) = -16(5)^2 + 32(5) + 40}}}


{{{s(5) = -400 + 160 + 40 = -200}}} which means that the rock is 200 feet <i>below</i> the ground.  An absurd result to be sure.  Double check the problem you were given because I suspect that the question was to determine the height at 0.5 seconds instead of the 5.0 seconds that you stated.


The second part of your problem needs you to recognize that the height function is a parabola and that it is concave down because the lead coefficient is negative.  That means the vertex of the parabola is a maximum.


For any parabola of the form {{{f(x) = ax^2 + bx + c}}}, the x-coordinate of the vertex is at {{{x = -b/2a}}}.


For your problem, {{{a = -16}}} and {{{b = 32}}}, so the maximum height is reached at time {{{-32/2(-16) = 1}}} second.


Calculate {{{s(1) = -16(1)^2 + 32(1) + 40}}} to get the actual maximum height reached.