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oksano4ka [1.4K]
1 year ago
13

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
Mathematics
1 answer:
klasskru [66]1 year ago
7 0

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:

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

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:

0=-16t^{2}+124t+19

Solving for t we get:

\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}

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

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Equations are written as two expressions joined by an equal sign. The terms on either side of the equal sign are called the "left side" and "right side" of the expression. It is often assumed that the right side of the equation is zero.

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