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?
hr height red = 96 t - 16 t^2
hg height green = 80 t -16 t^2
D = 96 t - 80 t = 16 t
b)32 ft
c) when the first one hits the ground.
D = 96 t - 80 t
Thank you very much.
a) 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.
The height of the red ball is given by the equation -16t^2 + 96t, and the height of the green ball is given by -16t^2 + 80t. Subtracting the green ball's height from the red ball's height, we have:
D(t) = (-16t^2 + 96t) - (-16t^2 + 80t)
D(t) = -16t^2 + 96t + 16t^2 - 80t
D(t) = 16t
Therefore, the polynomial D(t) that represents the difference in heights of the two balls is D(t) = 16t.
b) To find how much higher the red ball is 2 seconds after the balls are tossed, we can substitute t = 2 into the polynomial D(t) = 16t:
D(2) = 16(2)
D(2) = 32
Therefore, the red ball is 32 feet higher than the green ball 2 seconds after they are tossed.
c) To find when the difference in heights stops increasing, we need to determine the time when the red and green balls reach their maximum heights.
The maximum height for both balls occurs at the vertex of their respective quadratic equations, which can be found using the formula t = -b / (2a), where a is the coefficient of t^2 and b is the coefficient of t.
For the red ball's height equation, -16t^2 + 96t, we have a = -16 and b = 96.
t = -96 / (2*(-16))
t = -96 / (-32)
t = 3
Therefore, the red ball reaches its maximum height at t = 3 seconds.
For the green ball's height equation, -16t^2 + 80t, we have a = -16 and b = 80.
t = -80 / (2*(-16))
t = -80 / (-32)
t = 2.5
Therefore, the green ball reaches its maximum height at t = 2.5 seconds.
The difference in heights stops increasing when the time is equal to the maximum time for either ball, in this case, when t = 3 seconds.