Question 194403
You can use the equation for the height of an object propelled upwards from an initial height ({{{h[0]}}}) with an initial velocity ({{{v[0]}}}):
{{{h(t) = (1/2)gt^2+v[0]t+h[0]}}} where g is the acceleration due to gravity, 32 ft/sec. sq. Making the appropriate substitutions:
{{{h(t) = -16t^2+8t+24}}} 
To find the time the diver hits the water means to find the time (t) at which his height (h) is zero.
1) {{{-16t^2+8t+24 = 0}}} Factor out -8.
{{{-8(2t^2-t-3) = 0}}} Now factor the trinomial.
{{{2t^2-t-3 = (2t-3)(t+1)}}}={{{0}}} So...
{{{2t-3) = 0}}} or {{{t+1 = 0}}} and...
{{{t = 3/2}}} or {{{t = -1}}} Discard the negative solution as the time will be a positive value.
{{{t = 3/2}}}
The diver will hit the water in 1.5 seconds.
His maximum height can be found by finding the coordinates of the vertex (a maximum) of the downward-opening parabola represented by the quadratic equation.
The t-value is given by:
{{{t = -b/2a}}} and a = -16, b = 8.
{{{t = (-8)/2(-16)}}}
{{{t = 1/4}}}seconds. Now substitute this value of t into equation 1) above and solve for h.
{{{h(1/4) = -16(1/4)^2+8(1/4)+24}}}
{{{h(1/4) = -16(1/16)+(8/4)+24}}}
{{{h(1/4) = -1+2+24}}}
{{{h(1/4) = 25}}}feet.
The diver attains a maximum height of 25 feet above the water.