SOLUTION: A red ball and a green ball are tossed into the air. The red ball is given an initial velocity of 96 feet per second and its height "t" seconds after it is tossed is -16t^2 + 96t f

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Question 207153: A red ball and a green ball are tossed into the air. The red ball is given an initial 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 an initial velocity of 80 feet per second and its height "t" seconds after being tossed is -16t^2 +80t feet.
Find a polynomial D(t) that represents the difference in the heights of the two balls.
How much higher is the red ball 2 seconds after the balls are tossed?
When does the difference in the heights stop increasing?
Answer by ankor@dixie-net.com(22740) (Show Source):
You can put this solution on YOUR website!
A red ball and a green ball are tossed into the air. The red ball is given an
initial 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 an initial velocity of 80 feet per second and its height
"t" seconds after being tossed is -16t^2 +80t feet.
:
Find a polynomial D(t) that represents the difference in the heights of the two balls.
D(t) = (-16t^2 + 96t) - (-16t^2 +80t)
:
Remove brackets
D(t) = -16t^2 + 96t + 16t^2 - 80t
:
combine like terms:
D(t) = 16t
:
How much higher is the red ball 2 seconds after the balls are tossed?
D(t) = 16(2)
D(t) = 32 ft difference after 2 seconds
:
When does the difference in the heights stop increasing?
:
Obviously, when one of the balls hits the ground
:
Find when the lowest ball hits the ground (h=0)
-16t^2 + 80t = 0
Factor out -16t
-16t(t - 5) = 0
t = 5 seconds when the difference stops increasing;
:
A graph illustrates this well