Answer: 171.65 seconds.
Explanation
Given
[tex]h=-0.025t^2+4t+50[/tex]Procedure
We are asked how long will it take riders to pass over the hill and reach ground level, algebraically meaning when does h = 0. By rewriting our equation we get:
[tex]0=-0.025t^2+4t+50[/tex]In this case, we have a function in the form ax² + bx + c = 0. These types of equations can be solved by using the General Quadratic Formula:
[tex]x=\frac{-b\pm\sqrt{b^2-4ac}}{2a}[/tex]where a, b and c represents the coefficients in the form of the equation ax² + bx + c =0. In our case:
• a = - 0.025
,• b = 4
,• c = 50
By replacing the values and simplifying we get:
[tex]x=\frac{-4\pm\sqrt{(4)^2-4(-0.025)(50)}}{2(-0.025)}[/tex][tex]x=\frac{-4\pm\sqrt{16+5}}{-0.05}[/tex][tex]x=\frac{-4\pm\sqrt{21}}{-0.05}[/tex]Now, we have two solutions:
[tex]x_1=\frac{-4+\sqrt{21}}{-0.05}=-11.65[/tex][tex]x_1=\frac{-4-\sqrt{21}}{-0.05}=171.65[/tex]As we cannot have a negative time, then it will take 171.65 seconds.
