The Chs baseball team was on the field and the batter popped the ball up. The equation b(t) = 80t - 16t+3.45 represents the height of the ball above the ground in feet as a function of time in seconds. How long will the catcher have to get in position to catch the ball before it hits the ground?
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Answer:
t = 0.1 ± √(53)/40i where i = √(-1)
Step-by-step explanation:
CASE (1)
Given
Height as the function of time = B(t) = 80t² -16t + 3.45 .....(1)
When ball hits the ground the height of ball will zero
So
putting B(t) = 0 in equation (1) we get
80t² - 16t + 3.45 = 0
80t² -16t + 69/20 = 0
1600t² -320t + 69 = 0
by using Quadratic Formula
t = 0.1 ± √(53)/40i where i is the complex number equal to √(-1).
Hence in this case t = 0.1 ± √(53)/40i
CASE (2)
When
B(t) = 80t -16t + 3.45 = 64t + 3.45
Then
Putting B(t) = 0 for zero Height we get
64t +3.45 = 0
t = -3.45 /64
So in this case t = -3.45 /64 which is negative number
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