You drop a ball from a height of 98 feet. At the same time, your friend throws a ball upward. The polynomials represent the heights (in feet) of the balls after t seconds. a. Before the balls reach the same height, write a polynomial that represents the distance between your ball and your friend’s ball after t seconds. The polynomial is . Question 2 b. The constant term is , indicating the distance between the two balls is feet when they begin. The coefficient of the linear term is , indicating the two balls become feet closer to each other each second.
Answers
Answer:
98 - ut
Step-by-step explanation:
You drop a ball from a height of 98 feet. At the same time, your friend throws a ball upward. The polynomials represent the heights (in feet) of the balls after t seconds. a. Before the balls reach the same height, write a polynomial that represents the distance between your ball and your friend’s ball after t seconds. The polynomial is . Question 2 b. The constant term is , indicating the distance between the two balls is feet when they begin. The coefficient of the linear term is , indicating the two balls become feet closer to each other each second.
S = ut + (1/2)at²
Distance covered by ball dropped after t sec
u = 0 dropped
a = g ( acceleration due to gravity)
S = (1/2) g t²
Height from Ground after t sec = 98 - S = 98 - gt²/2
Height from Ground of ball dropped after t sec = 98 - gt²/2
Ball thrown upward after t sec
S = ut - (1/2)gt²
u = initial speed by which ball thrown upward
Height of ball thrown upward after t sec = ut - gt²/2
Distance between Dropped & thrown upward ball = (98 - gt²/2) - (ut - gt²/2)
= 98 - ut
The constant term = 98
The coefficient of the linear term = u (speed by which ball is thrown upward)
Ball will reach at same height when 98 - ut = 0
=> t = 98/u
Answer:
98
Step-by-step explanation:
it is 98