Question 720469
The effect of gravity pulling down is {{{ -16t^2 }}} ft/sec2
The effect of upward initial velocity is {{{ 48t }}} ft/sec
The hight off the ground adds {{{ 8 }}} ft
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The equation is {{{ h = -16t^2 + 48t + 8 }}}
(a) The maximum height is at {{{t =  -b/(2a) }}} for
a parabola ( which this is ), when the form of the
equation is {{{ h = at^2 + b*t + c }}}
{{{ a = -16 }}}
{{{ b = 48 }}}
{{{ -b/(2a) = -48/(2*(-16)) }}}
{{{ -b/(2a) = 1.5 }}}
So, at t = 1.5 sec, the height is a maximum
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(b)
Plug {{{ t = 1.5 }}} back into the equation to find {{{h}}}
{{{ h[max] = -16*1.5^2 + 48*1.5 + 8 }}}
{{{ h[max] = -36 + 72 + 8 }}}
{{{ h[max] = 44 }}}
The max height is 44 ft
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(d)
In 3 sec, the ball will be
{{{ h = -16t^2 + 48t + 8 }}}
{{{ h = -16*3^2 + 48*3 + 8 }}}
{{{ h = -144 + 144 + 8 }}}
In 3 sec, the ball will be 8 ft off the ground again
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(e)
The axis of symmetry is a vertical line, and every 
point on that line has the same x-value, so
the equation for this line is {{{ x = 1.5 }}}
Here's the plot of the equation:
{{{ graph( 400, 400, -1, 5, -5, 50, -16x^2 + 48x + 8 ) }}}