Question 205029
A red and green ball are tossed in the air at the same time. The red ball is
 given the velocity of 96 feet per second and its height (t) seconds after it
 is tossed is -16t^2+96t feet. The green ball is given a velocity of 80t feet
 per second and its height (t) seconds after it is tossed is -16t^2+80t feet.
 What is the polynomial difference (t) that represent the difference in the
heights of the two balls? How much higher is the read ball 2 seconds after the
 balls are tossed? In reality, when does the difference in the heights stop
 increasing?
 Do I subtract the second polynomial term from the first to get the difference?
:
Yes, indeed, that's what you do:
f(t) = (-16t^2 + 96t) - (-16t^2 + 80t)
Remove bracket, change signs after the negative;
f(t) = -16t^2 + 96t + 16t^2 - 80t
:
f(t) = -16t^2 + 16t^2 + 96t - 80t
:
f(t) = 16t; the difference between the height of the two balls
;
How much higher is the red ball 2 seconds after the  balls are tossed?
t = 2
f(t) = 16(2)
f(t) = 32 ft higher is the red ball
:
 In reality, when does the difference in the heights stop  increasing?
:
When the green ball hits the ground height, h = 0
Find the time when the height of the green ball = 0
-16t^2 + 80t = 0
Factor out -16t
-16t(t - 5) = 0
the solution we want is 
t = 5 seconds
:
:
A graph will illustrate this nicely
:
{{{ graph( 300, 200, -4, 8, -20, 150, -16x^2+96x, -16x^2+80x) }}}
Note the green ball strikes the ground after 5 sec, while the red ball is still in the air, After 6 sec, both are on the ground and the difference is 0