You throw a ball up and its height h can be tracked using the equation: h = 2x^2 − 12x + 20. What is the highest the ball will reach? B) When will the ball land (i.e. hit 0 feet high)?
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At 3 Sec , Ball will reach highest 2 feet
Step-by-step explanation:
Correction
h = 2x² - 12x + 20
dh/dt = 4x - 12
Velocity will be zero at highest point
=> 4x - 12 = 0
=> x = 12/4
=> x = 3
After 3 Secs Ball will reach highest point
Height at 3 secs
= 2(3)² - 12(3) + 20
= 18 - 36 + 20
= 38 - 36
= 2
At 3 Sec , Ball will reach highest 2 feet
When ball will land
h = 0
2x² - 12x + 20 = 0
=>x² - 6x + 10 = 0
x = (6 ± √-4)/2
= ( 6 ± 2i)/2
= 3 ± i
theoretically 3 + i will be time when ball will land
this is an imaginary number
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