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An object is thrown upward at a speed of 124 feet per second by a machine from a height of 19 feet off the ground. The height h of the object after t seconds can be found using the equation h = - 16t2 + 124t + 19 When will the height be 59 feet? Select an answer When will the object reach the ground? Select an answer > Next Question

Sagot :

a) For this question we set h=59 and solve for t, in order to do so we use the general formula for second-degree equations:

[tex]\begin{gathered} t=\frac{-124\pm\sqrt[]{124^2-4(-16)(-40)}}{2(-16)} \\ t=\frac{-124\pm113.21}{-32} \end{gathered}[/tex]

The height of the object will be 59 feet at t=7.41 seconds and t=0.34 seconds.

b) When the object reaches the ground, h=0 therefore:

[tex]0=-16t^{2}+124t+19[/tex]

Solving for t we get:

[tex]\begin{gathered} t=\frac{-124\pm\sqrt[]{124^2-4(-16)(19)}}{2(-16)} \\ t=\frac{-124\pm\sqrt[]{16592}}{-32}=\frac{-124\pm128.81}{-32} \end{gathered}[/tex]

Therefore, since t cannot be negative the solution is t=7.9 seconds.

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