Question 730246
{{{ h(t) = -16t^2 + 48t + 32 }}}
(a)
The instant the ball is thrown upward is
called {{{ t = 0 }}}. Plug this value into the equation.
{{{ h(0) }}} is the height of the building.
{{{ h(0) = -16*0 + 48*0 + 32 }}}
{{{ h(0) = 32 }}} ft
(b)
The path of the ball is a parabola. The t-coordinate
of the vertex is at {{{ t[v] = -b/(2a) }}} when the form
is like the given equation
{{{ a = -16 }}}
{{{ b = 48 }}}
{{{ -b/(2a) = -48/( 2*(-16)) }}}
{{{  t[v] = 3/2 }}}
Plug this back into equation to find {{{ h[v] }}}
{{{ h[v] = -16*(3/2)^2 + 48*(3/2) + 32 }}}
{{{ h[v] = -36 + 72 + 32 }}}
{{{ h[v] = 68 }}}
The max height is 68 ft
(c)
The height is zero when it hits the ground, so
{{{ 0 = -16t^2 + 48t + 32 }}}
{{{ -t^2 + 3t + 2 = 0 }}}
Use the quadratic formula
{{{ t = (-b +- sqrt( b^2-4*a*c ))/(2*a) }}} 
{{{ a = -1 }}}
{{{ b = 3 }}}
{{{ c = 2 }}}
{{{ t = (-3 +- sqrt( 3^2-4*(-1)*2 ))/(2*(-1)) }}} 
{{{ t = (-3 +- sqrt( 17 ))/(-2)) }}} 
{{{ t = ( 3 + sqrt(17)) / 2 }}}
{{{ t = ( 3 + 4.123 ) / 2 }}}
{{{ t = 3.562 }}} 
The ball hits the ground in 3.562 sec
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Here's the plot of the equation:
{{{ graph( 400, 400, -1, 5, -5, 75, -16x^2 + 48x + 32) }}}