<PRE><B>Question 192982</B>
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Two sides and a diagonal of a square form an isosceles right triangle.  If the length of the side is &lt;b&gt;&lt;i&gt;s&lt;/i&gt;&lt;/b&gt; and the length of the diagonal is &lt;b&gt;&lt;i&gt;s&lt;/i&gt; + 2&lt;/b&gt;, then Pythagoras&#39; theorem says:


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ (s + 2)^2 = 2s^2]


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ s^2 + 4s + 4 - 2s^2 = 0]


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ s^2 - 4s - 4 = 0]


Solve the quadratic for &lt;b&gt;&lt;i&gt;s&lt;/i&gt;&lt;/b&gt;.  The negative root is extraneous and your answer is the positive root.  I recommend that you leave your answer in radical form.


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*[tex \LARGE \ \ \ \ \ \ \ \ \ \ S(t) = -16t^2 + v_ot + s_o]


Substitute the given values for initial velocity and initial height, then set the function equal to zero, because zero is the height when the ball hits the ground, then solve the quadratic for &lt;b&gt;&lt;i&gt;t&lt;/i&gt;&lt;/b&gt;.


Again, exclude any negative root -- the ball can&#39;t possibly return to the ground before you threw it.


John
*[tex \LARGE e^{i\pi} + 1 = 0]
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