Question 150370
To start solving this type of problem, you should know that the height (h) of an object propelled upward from a height of {{{h[0]}}} from the earth's surface with an initial velocity of {{{v[0]}}}ft./sec. is given by the function:
{{{h(t) = -16t^2+v[0]t+h[0]}}}
In this problem, {{{v[0] = 40}}}ft./sec. and {{{h[0] = 30}}}ft. and you want to find the height, h, when t = 2 secs., so, substitute t = 2 and...
{{{h(2) = -16(2)^2+40(2)+30}}}
{{{h(2) = -64+80+30}}}
{{{h(2) = 46}}}
1) The rock is {{{highlight(46)}}} feet high after 2 seconds.
The maximum height can easily be found when you realize that this quadratic function, when graphed, represents a parabola that opens downwards (negative {{{x^2}}} coefficient), so you need to find the location of the vertex (maximum) of the parabola.
The independent variable is t (time) so the t-coordinate of the vertex is given by:
{{{t = -b/2a}}} where the a and b come from the standard form of the quadratic equation:{{{y = ax^2+bx+c}}}, so, in this problem,  a = -16 and b = 40
{{{t = (-40)/2(-16)}}}
{{{t = 40/32}}}
{{{t = 1.25}}}secs.
The maximum height occurs at time {{{highlight(t = 1.25)}}} seconds. Now substitute this value of t into the quadratic function to find the height of the rock at time t = 1.25.
{{{h(1.25) = -16(1.25)^2+40(1.25)+30}}}
{{{h(1.25) = 55}}}feet.
The maximum height attained by the rock is {{{highlight(h = 55)}}} feet.