Question 139518
Let's look at part I of the problem:
I)
a) How many seconds for the ball to go up and down?
In other words, how many seconds for the ball to return to the ground?
Starting with the given formula:
{{{h(t) = -16t^2+v[0]t+h[0]}}} where: 
{{{v[0] = 132}}}ft/sec This is the initial upward velocity.
{{{h[0] = 0}}}  This is the initial height.
So we have:
{{{h(t) = -16t^2+132t}}} and you want to find the time, t, when h = 0: 
{{{0 = -16t^2+132t}}} Factor a t from the right-hand side.
{{{0 = t(-16t+132)}}} Apply the zero product rule: If {{{a*b = 0}}} then either {{{a = 0}}} or {{{b = 0}}} or both. 
Add 16t to both sides. So...
{{{t = 0}}} or {{{-16t+132 = 0}}}
{{{-16t + 132 = 0}}} Add 16t to both sides.
{{{132 = 16t}}} Divide both sides by 16.
{{{t = 8.25}}}seconds.
The ball returns to the ground in 8.25 seconds.
b) How high does the ball go?
You'll need to find the value of t at the vertex of the parabola described by the given equation.  This is given by:
{{{t = -b/2a}}} The a and b come from: {{{ax^2+bx+c = 0}}} and in this case, a = -16 and b = 132.
{{{t = (-132)/2(-16)}}}
{{{t = 4.125}}} seconds.  This is the time at which the ball reaches its maximum height.  Substitute this value of t into the given equation and solve for h to find the maximum height.
{{{h(4.125) = -16(4.125)^2+132(4.125)}}}
{{{h(4.125) = -272.5+544.5}}}
{{{h(4.125) = 272.25}}} feet. Which is precisely what you got.
II) For this problem to be solved for the planet Mercury whose gravitational force is 38% that of earth's, let's look at the original formula for the height of an object propelled upwards:
{{{h(t) - (1/2)gt^2+v[0]t+h[0]}}}
Here, the acceleration due to gravity is: g = 32 ft/second squared for earth.
For Mercury, g = 0.38(32)ft/second squared = 12.16 feet/second squared.
So the formula for the planet Mercury becomes:
{{{h(t) = -(1/2)12.16t^2+v[0]+h[0]}}}
{{{h(t) = -6.08t^2+v[0]t+h[0]}}} This is the formula you would use to solve the second part of the problem.
a)
{{{h(t) = -6.08t^2+132t}}} Set h = 0 and solve for t.
{{{0 = -6.08t^2+132t}}} So...
{{{t = 0}}} or...
{{{t = 132/6.08}}}
{{{t = 21.71}}} seconds.  This is the time at which the ball returns to the ground on Mercury.
The time to reach its maximum height is:
{{{t = -b/2a}}} where: a = -6.06 and b = 132.
{{{t = -(132/2(-6.08))}}}
{{{t = 10.86}}}seconds. Substitute this into the formula for Mercury:
{{{h(10.86) = -6.08(10.86)^2+132(10.86)}}}
{{{h(10.86) = -66+1433.52}}}
{{{h(10.86) = 1367.52}}}feet. This is the maximum height attained by the ball on Mercury.