Aprojectile is fired straight up from ground level with an initial velocity of 112 ft/s. its height, h, above the ground after t seconds is given by h = –16t^2 + 112t. what is the interval of time during which the projectile's height exceeds 192 feet?

When shot, the projectile will go straight up, but only for some while.  
All the way up, it'll keep going slower and slower until it completely runs 
out of speed.  Then, at the highest point, it'll start falling, and continue 
falling faster and faster until it hits the ground.
So ... for any height you choose (less than the peak height), the projectile 
will be at that height twice: once on the way up, and again on the way down.
All we have to do is find the two times when the projectile is at 192 feet, and we'll know that it's higher than 192 feet between those two times.
Fortunately, we have a second-degree equation for the height, so we know that the equation will have two solutions for any height less than the peak.  

The Golden Equation:            H  =  -16 t²  + 112 t

We want to know:  What time
is 't' when 'H' is 192' ?              192  =  -16 t²  + 112 t

Subtract 192 from each side:      -16 t² + 112 t - 192  =  0

(Just to make it feel better,
multiply each side by -1 ):             16 t² - 112 t + 192  =  0  

Divide each side by  16 :                  t² -    7  t  + 12   =  0

Stuff this equation through
the quadratic formula, and
we discover that                            t = 3
                                             and
                                                   t = 4 .
This just has the sweet feel 
of success all over it.

The projectile is 192 ft above the ground at 3 seconds, on the way up, 
and again at 4 seconds, on the way down.  Between 3 and 4 seconds, 
it's higher than that.

Choice-a is    3 < t < 4 .

Answer ( slope);