Question 1076210
{{{ h = -16t^2 + 80t }}}
(a)
The ball is on the ground ( h = 0 ) at 2 times
(1) instant the ball is hit ( t = 0 )
(2) instant the ball returns to ground
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Set the height equal to zero
{{{ 0 = -16t^2 + 80t }}}
{{{ t*( -16t + 80 ) = 0 }}}
{{{ -16t + 80 = 0 }}}
{{{ 16t = 80 }}}
{{{ t = 5 }}}
The ball hits the ground in 5 sec
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(b)
The t-value of the maximum height is given
by the formula:
{{{ t[max] = -b/(2a) }}}
when the equation looks like:
{{{ h = a*t^2 + b*t + c }}} ( note that {{{c=0}}} )
{{{ a = -16 }}}
{{{ b = 80 }}}
{{{ t[max] = -80/(2(-16)) }}}
{{{ t[max] = 80/32 }}}
{{{ t[max] = 2.5 }}} sec
Plug this result back into equation
{{{ h[max] = -16t^2 + 80t }}}
{{{ h[max] = -16*2.5^2 + 80*2.5 }}}
{{{ h[max] = -16*6.25 + 200 }}}
{{{ h[max] = -100 + 200 }}}
{{{ h[max] = 100 }}} 
The maximum height is 100 ft
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Here's the graph:
{{{ graph( 400, 400, -1, 6, -10, 120, -16x^2 + 80x ) }}}