Question 11365
Well, I'm not surprised to hear that you have been struggling with this problem. You seem to missing the description of the golf ball's behavior following its initial (upward?) velocity from the ground.

The formula is:

{{{h(t) = -4.9t^2 + (Vo)t + Ho}}} This function describes the height (in meters) of an object propelled upward with an initial velocity of Vo (meter per second) from an initial height of Ho (meters).

a) {{{h(t) = -4.9t^2 + 4.9t}}}  Since the golf ball is leaving from the ground, the initial height (Ho) is zero.

b) When the golf ball will returns to the ground, its height (h) will be zero.

{{{0 = -4.9t2 + 4.9t}}} Solve for t.

{{{t(-4.9t + 4.9) = 0}}}

{{{t = 0}}} or {{{-4.9t + 4.9 = 0}}}

t = 0 is when the golf ball started it epic journey into spase.

{{{-4.9t = -4.9}}}
{{{t = 1}}} The ball returns to Earth in 1 second.

c) To find when the ball reaches its zenith (highest point), you need to find the time, t, at the vertex of the parabola. 
This can be found by using t = -b/2a (Assuming that you don't use differential calculus). 

{{{t = (-4.9)/2(-4.9)}}}
{{{t = 1/2}}}

The golf ball reaches its maximum height in 1/2 second.

d) Substitute t = 1/2 second into the original function, h(t) to find its maximum height.

{{{h(1/2) = -4.9(1/2)^2 + 4.9(1/2)}}} Solve for h.
{{{h(1/2) = (-4.9/4) + 4.9/2}}}
{{{h(1/2) = -4.9/4 + 9.8/4}}}
{{{h(1/2) = 4.9/4}}}
{{{h(1/2) = 1.22}}}

The maximum height attained by the golf ball is 1.22... meters.