Question 386723
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You have a couple of basic problems with the function as you presented it.  In the first place, your lead coefficient is wrong.  9.8 m/sec^2 is the acceleration due to gravity but the general height function for vertical motion near the surface of planet Earth is:


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ h(t)\ =\ \frac{1}{2}gt^2\ +\ v_ot\ +\ h_o]


Where *[tex \Large g\ =\ -9.8\text{ m/sec^2}] and is the acceleration due to gravity (minus because gravity accelerates objects downward), *[tex \Large v_o] is the initial vertical velocity component, and *[tex \Large h_o] is the initial height.


Your function also presupposes a 10 m/sec downward (hence the minus sign) vertical component to the initial velocity component.


So your function is actually:


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ h(t)\ =\ -4.9t^2\ -\ 10t\ +\ 310]


You want to know the time it will take for the stone to hit the ground.  Not surprisingly (I hope) the ground is 0 feet.  So we make *[tex \Large h(t)\ =\ 0] and solve for t.


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ -4.9t^2\ -\ 10t\ +\ 310\ =\ 0]


Looks an awful lot like a quadratic equation to me.  Use the quadratic formula.  We know that we will have a pair of real number roots.  Any time your quadratic has opposite signs on the lead and constant coefficients, you always have real number roots.  One of the roots will be a negative number which you should discard since you can't go back in time.  The positive root is the number of seconds it will take for the stone to hit the ground.


Here's the quadratic formula in case you have forgotten:


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ x\ =\ \frac{-b\ \pm\ \sqrt{b^2\ -\ 4ac}}{2a}]


John
*[tex \LARGE e^{i\pi} + 1 = 0]
My calculator said it, I believe it, that settles it
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