Height difference. A red ball and a green ball are
simultaneously 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 �16t2 � 96t feet. The green ball
is given an initial velocity of 80 feet per second, and its
height t seconds after it is tossed is �16t2 � 80t feet.
a) Find a polynomial D(t) that represents the difference in
the heights of the two balls.
b) How much higher is the red ball 2 seconds after the
balls are tossed?
c) In reality, when does the difference in the heights stop
increasing?
a) To find the difference in heights of the two balls, we subtract the height of the green ball from the height of the red ball.
Height of red ball = -16t^2 + 96t
Height of green ball = -16t^2 + 80t
Therefore, the polynomial D(t) representing the difference in heights is:
D(t) = (-16t^2 + 96t) - (-16t^2 + 80t)
= 16t
b) To find how much higher the red ball is 2 seconds after the balls are tossed, we substitute t = 2 into the polynomial D(t):
D(2) = 16(2)
= 32 feet
The red ball is 32 feet higher 2 seconds after the balls are tossed.
c) To find when the difference in heights stops increasing, we need to determine the value of t where D(t) reaches its maximum. Since D(t) = 16t is a linear function, it increases indefinitely without a maximum value.
Therefore, the difference in heights never stops increasing.
To find the polynomial D(t) that represents the difference in the heights of the two balls, we need to subtract the height of the green ball from the height of the red ball.
Given that the height of the red ball is given by the equation -16t^2 + 96t feet, and the height of the green ball is given by the equation -16t^2 + 80t feet, we can subtract the two equations as follows:
D(t) = (-16t^2 + 96t) - (-16t^2 + 80t)
= -16t^2 + 96t + 16t^2 - 80t
= 16t
Therefore, the polynomial D(t) that represents the difference in the heights of the two balls is D(t) = 16t.
To find how much higher the red ball is 2 seconds after the balls are tossed, we need to substitute t = 2 into the equation for the height of the red ball:
Height of red ball at t = 2 seconds = -16(2)^2 + 96(2)
= -64 + 192
= 128 feet
Similarly, we substitute t = 2 into the equation for the height of the green ball:
Height of green ball at t = 2 seconds = -16(2)^2 + 80(2)
= -64 + 160
= 96 feet
The difference in height 2 seconds after the balls are tossed is:
D(2) = 128 - 96
= 32 feet
Therefore, the red ball is 32 feet higher than the green ball 2 seconds after the balls are tossed.
To find when the difference in heights stops increasing, we need to determine when the velocity of the red ball becomes equal to the velocity of the green ball.
The velocity of the red ball is given by the derivative of its height equation: v(t) = -32t + 96.
The velocity of the green ball is given by the derivative of its height equation: v(t) = -32t + 80.
Setting these two equations equal, we get:
-32t + 96 = -32t + 80
Simplifying this equation, we find:
16 = 0
Since the equation is not true, there is no value of t for which the velocities are equal. Therefore, the difference in heights continues to increase indefinitely, and there is no time when it stops increasing.