Answer:
Third option: 5.48
Step-by-step explanation:
Given the quadratic model that represents the approximate height in meters of the ball [tex]f(x)=-5x^2 +200[/tex]. the time in seconds x, when the ball is 50 meters from the ground can be calculated by substituting [tex]f(x)=50[/tex] and solving for x.
Then:
[tex]50=-5x^2 +200[/tex]
Subtract 200 from both sides:
[tex]-200+50=-5x^2 +200-200\\-150=-5x^2[/tex]
Divide both sides by -5:
[tex]\frac{-150}{-5}=\frac{-5x^2}{-5}\\30=x^2[/tex]
Apply square root to both sides:
[tex]\sqrt{30}=\sqrt{x^2}\\\sqrt{30}=x\\5.48=x[/tex]
The ball is 50 meters from the ground after about 5.48 seconds.
