Question 182182
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The height function for an object propelled upward at a given initial vertical velocity in feet per second from a given initial height in feet is:


*[tex \LARGE \text{          }\math h(t) = -16t^2 + v_ot + s_o]


where *[tex \Large s_o] is the initial height, *[tex \Large v_o] is the initial velocity, and -16 is the acceleration due to gravity.  In your case:


*[tex \LARGE \text{          }\math h(t) = -16t^2 + 150t + 5]


You want to solve for the time, <i>t</i>, that it will take <i>h(t)</i> to reach 0, so:


*[tex \LARGE \text{          }\math -16t^2 + 150t + 5 = 0]


You had the right idea, except for two things:  First you did not account properly for the acceleration due to gravity.  Your acceleration factor was +1 rather than -16.  According to the way you were modeling the situation, the ball would keep going up forever, and keep getting faster as it was going.  Secondly, and not nearly as important as having the proper acceleration, you had the constant term as -5.  That means that Mr. O'Brian was able to strike the ball 5 feet <i><b>below</b></i> the surface of the ground.


Aside from all of that, you had the quadratic formula correctly expressed.  For any 2nd degree polynomial equation of the form: 


*[tex \LARGE \text{          }\math ax^2 + bx + c = 0],


The two roots of the equation are given by:


*[tex \LARGE \text{          }\math x_{1,2} = \frac {-b \pm sqrt{b^2-4ac}}{2a}]


All you have to do is plug in the numbers:


*[tex \LARGE \text{          }\math t_{1,2} = \frac {-(150) \pm sqrt{(150)^2-4(-16)(5)}}{2(-16)}]


And do the arithmetic.


Note that you will arrive at two solutions.  Given correct arithmetic, you will get a very small magnitude negative number and a larger magnitude positive number.  The small negative number represents the time it would have taken the ball to get from ground level (i.e. 0 feet) to 5 feet if the ball had started there.  The positive root is the solution you seek.




John
*[tex \LARGE e^{i\pi} + 1 = 0]
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