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Sagot :
Answer:
[tex]9.99\text{secs}[/tex]Explanation:
Here, we want to find the time the rocket will hit the ground
What we have to do is to substitute 0 for the height of the rocket y and solve the quadratic equation that results
That simply means we are solving for:
[tex]0=-16x^2+149x+108[/tex]We can use the quadratic formula to solve this
[tex]x\text{ = }\frac{-b\pm\sqrt[]{b^2-4ac}}{2a}[/tex]where a is the coefficient of x^2 which is -16
b is the coefficient of x which is 149
c is the last number which is 108
Substituting the values, we have it that:
[tex]\begin{gathered} x\text{ = }\frac{-149\pm\sqrt[]{149^2-4(-16)(108)}}{2(-16)} \\ \\ x\text{ = }\frac{-149\pm\sqrt[]{29113}}{-32}\text{ } \\ \\ x\text{ = }\frac{-149-170.625\text{ }}{-32} \\ or\text{ } \\ x\text{ = }\frac{-149_{}+170.625}{-32} \end{gathered}[/tex]Now, we proceed to get the individual x-values:
[tex]\begin{gathered} x\text{ = }\frac{-149-170.625}{-32}\text{ = 9.99 } \\ or \\ x\text{ = }\frac{-149+170.625}{-32}\text{ = -0.68} \end{gathered}[/tex]Since time cannot be negative, we use the first value only
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