Question 607060
The minus sign in front of the {{{t^2}}}
term tells me that this 2nd degree
curve has a maximum and not a minimum.
The t-coordinate of the vertex ( maximum )
is at {{{ t[v] = -b/(2a) }}} when the 
form is {{{ at^2 + bt + c }}}
{{{ a = -4.9 }}}
{{{ b = 4 }}}
{{{ t[v] = -4/(2*(-4.9)) }}}
{{{ t[v] = 2/4.9 }}}
{{{ t[v] = .4082 }}}
Plug this back into equation to find {{{ h(.4082) }}}
{{{ h(t) = -4.9t^2 + 4t + 8 }}}
{{{ h(.4082) = -4.9*(.4082)^2 + 4*.4082 + 8 }}}
{{{ h(.4082) = -4.9*.1666 + 1.6326 + 8 }}}
{{{ h(.4082) = -.8163 + 1.6326 + 8 }}}
{{{ h(.4082) = -.8163 + 1.6326 + 8 }}}
{{{ h(.4082) = 8.8163 }}}
The greatest height in cm is 8.8163 cm
here's a plot:
( ignore the negative values of {{{t}}} )
{{{ graph( 400, 400, -2, 4, -2, 10, -4.9x^2 + 4x + 8 ) }}}