Questions on Word Problems: Travel and Distance answered by real tutors!

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Question 150358: Height difference
A red ball and 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 -16t^2+p6ft. The green ball is given an initial velocity of 80 feet per second and its height t seconds after it is tossed is -16t^2 + 80 ft
a. find the polynomial that represents the difference in the heights of the two balls.
b. how much higher is the red ball 2 seconds after the ball is tossed?
c. in reality when does the difference in the heights stop increasing?
: Height difference
A red ball and 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 -16t^2+p6ft. The green ball is given an initial velocity of 80 feet per second and its height t seconds after it is tossed is -16t^2 + 80 ft
a. find the polynomial that represents the difference in the heights of the two balls.
b. how much higher is the red ball 2 seconds after the ball is tossed?
c. in reality when does the difference in the heights stop increasing?

Answer by ankor@dixie-net.com(4189) About Me  (Show Source):
You can put this solution on YOUR website!
A red ball and 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 -16t^2+p6ft. The green ball is given an initial velocity of 80 feet per second and its height t seconds after it is tossed is -16t^2 + 80 ft
:
a. find the polynomial that represents the difference in the heights of the two balls.
Difference = D(t)
D(t) = (-16t^2 + 96t) - (-16t^2 + 80t)
D(t) = -16t^2 + 96t + 16t^2 - 80t
D(t) = 16t
:
b. how much higher is the red ball 2 seconds after the ball is tossed?
D(t) = 16(2)
D(t) = 32 ft
:
c. in reality when does the difference in the heights stop increasing?
:
After 5 sec, the green ball hits the ground: h(t)= -16(25) + 80(5) = 0