Determine the height of the Mountain.... 1 Attachment(s) I am doing number 2 a is the height of the moutian as easy as just the 500? Thanks 
yes ... height of the mountain is h(0) = 500 meters 
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$a$ is the coefficient of the quadratic term, $a = 4.9$ $b$ is the coefficient of the linear term, $b = 196$ $c$ is the constant term, $c=500$ 
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Okay, I think I messed up my words again... I guess I am just confused how they got the numbers? How did they know what numbers to put in for a and c? I am guessing it's because they already had the numbers? Cause now they want me to plug in numbers for a b and c I think, but by writing a new.... Assuming its height is a quadratic function of time, how long will it spend in the air? A ball is thrown into the air from 6 feet above the ground. It reaches its maximum height of 18 feet in 3 seconds So I would think c=6 18 is maybe our b and a is maybe our unknown? 
The general equation for projectile motion in one dimension is $h(t) = \dfrac{1}{2}gt^2 + v_0 \cdot t + h_0$ in English units, $g = 32 \, ft/sec^2$ in metric units, $g = 9.8 \, m/sec^2$ Quote:
$a = 16$ The $b$ value is the initial velocity of the projectile ... $b = v_0$ The $c$ value is the initial height of the projectile ... $c = h_0$ You are given $h(0) = 6 = c$ and $h(3) = 18$ ... $18 = 16(3^2) + v_0(3) + 6$ Now you can solve for $b = v_0$ and determine the equation for height as a function of time. 
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Hmm okay, so other than from you, where would I have learned that g equals 32? and to halve it as you have? 
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(a) your textbook if it has examples of these projectile motion problems (b) your instructor if he/she did example problems in class (c) yourself, if you did a web search ... Quadratics in the real world 
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going back to the original problem /application 1 Attachment(s) How come I am not coming up with a Decimal ? Thanks 
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