<PRE><B>Question 708281</B>
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(a)  Set the function equal to 240.


*[tex \LARGE -16t^2\ +\ v_ot\ =\ 240]


Substitute your given value of *[tex \LARGE v_o] and then solve the quadratic.  The equation will have two zeros, one representing the time the projectile reaches 240 feet on the way up, and the other representing the time the projectile passes 240 feet again on the way back down.


(b)  Same as part (a) except you set the function equal to the height of the ground, i.e. zero.  Your two zeros will be zero and some positive number.  The zero root represents the instant of launch, and the other is the time of return to ground.


John
*[tex \LARGE e^{i\pi}\ +\ 1\ =\ 0]
&lt;font face=&quot;Math1&quot; size=&quot;+2&quot;&gt;Egw to Beta kai to Sigma&lt;/font&gt;
My calculator said it, I believe it, that settles it
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