Question 1038747
Start with the general form of a parabola
which has a maximum and not a minimum
{{{ h = -a*t^2 + b*t + c }}} ( {{{ a }}} must be positive )
Each point on the parabola is  ( t,h )
----------------------------
Since the ball is dropped and not thrown upward,
The vertex ( peak ) is at {{{ t[v]=0 }}}
The formula for the vertex is:
{{{ t[v] = -b/(2a) }}}
{{{ 0 = -b/(2a) }}}, so {{{ b=0 }}}
-----------------------------
The parabola must have the form:
{{{ h = -a*t^2 + c }}}
They give you 2 points:
( 3, 432 )
( 6, 0 )
Use these points to find {{{ a }}} and {{{ c }}}
----------------
{{{ 432 = -a*3^2 + c }}}
(1) {{{ -9a + c = 432 }}}
and
{{{ 0 = -a*6^2 + c }}}
(2) {{{ -36a + c = 0 }}}
-----------------------
Subtract (2) from (1)
(1) {{{ -9a + c = 432 }}}
(2) {{{ 36a - c = 0 }}}
-----------------------
{{{ 27a = 432 }}}
{{{ a = 16 }}}
and
(2) {{{ -36a + c = 0 }}}
(2) {{{ -36*16 + c = 0 }}}
(2) {{{ c = 576 }}}
---------------------
So, the equation is:
{{{ h = -16t^2 + 576 }}}
Here's the plot:
{{{ graph( 400, 400, -1, 8, -70, 700, -16x^2 + 576 ) }}}