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what would the standard form be for this question? During a drumline performance, a drummer throws his drumstick with an upward velocity of 32 feet per second. if the drummer releases and catches the drumstick 6 feet above the ground, what is the maximum height of the drumstick? how long is the drumstick in the air? what i cant figure out is what the ax^2 is i can get the bx and the c but not the A. please help.

mark.H
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2 Answers2

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Your Equation for this kind of problem is

$s(t) = –gt^2 + v_0t + h_0$

$g$ is Gravity (half gravity) you were given a velocity in terms of feet per second so we will go with $16$ for this value

$v$ is initial velocity which you were given as $32$

$h$ is your initial height which is $6$ ft

So your equation is

$s(t) = -16t^2 + 32t + 6$

Malachi
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  • What is t? and, shouldn't the equation used be $s = 6 + \frac{-(v^2)}{2 a}$ where $v$ is the initial velocity (32ft/s) and $a$ is gravitational acceleration ( $-32.174fts^{-2}$ ) – rolfl Dec 12 '14 at 16:01
  • I thought I left a Link to where I got the equation, and the values. $t$ is the $x$ variable – Malachi Dec 12 '14 at 16:04
  • $t$ is the $x$ variable or Time in Seconds. so we get a parabola to indicate the height of the drumstick over time – Malachi Dec 12 '14 at 16:41
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Hint: Assume the height function is $$f(t) = at^2 + bt + c\tag{1}$$ Then you have three initial conditions: $$f''(0) = -32 \tag{2}$$ $$f'(0)=32\tag{3}$$ $$f(0)=6\tag{4}$$

These conditions should allow you to determine the unknown coefficients of $f$.

MPW
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