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Chapter 11: Problem 106

$$\text { Solve for } t: \quad s=-16 t^{2}+v_{0} t$$

Short Answer

Expert verified
The correct solutions for \(t\) from the given quadratic equation are \[\frac{-v_{0} \pm \sqrt{v_{0}^{2} +64s}}{-32}\]

Step by step solution

01

Rearrange in standard quadratic form

First, the equation should be rearranged to match the standard form for a quadratic equation, which is usually written as \(at^{2} + bt + c = 0\). The given equation is already in this form once we consider \(a = -16\), \(b = v_{0}\), and \(c = -s\). Then the equation becomes \(-16t^{2} + v_{0}t + s = 0.\)
02

Use the quadratic formula

With the equation set up, apply the quadratic formula, which is \(-b \pm \sqrt{b^{2}-4ac}\) over \(2a\). Inserting \(a = -16\), \(b = v_{0}\), and \(c = s\) into the formula, we get \(t = \frac{-v_{0} \pm \sqrt{v_{0}^{2} - 4*(-16)*s}}{2*(-16)}\). Simplify all under the square root first, then divide by \(-32\).
03

Simplify

First, compute underneath the square root: \(v_{0}^{2} +64s\). Divide the result and also \(-v_{0}\) by \(-32\) to reveal the solutions for \(t\).

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