Question #5873d

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Question #5873d

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Answer 1 · 2018-02-15T09:29:15

At $0.8 s$ $0.8s$ and $3.6 s$ $3.6s$ , the stone will be at a height of $15 m$ $15m$ .

Explanation:

Since the formula for $h (t)$ $h(t)$ is $22 t - 5 t^{2}$ $22t-5t^2$ , and here, $h (t) = 15$ $h(t)=15$ , we can write:

$22 t - 5 t^{2} = 15$ $22t-5t^2=15$

$22 t - 5 t^{2} - 15 = 0$ $22t-5t^2-15=0$

$5 t^{2} - 22 t + 15 = 0$ $5t^2-22t+15=0$

Now use the quadratic formula to solve for $t$ $t$ .

For any $a x^{2} + b x + c = 0$ $ax^2+bx+c=0$ , where $a \neq 0$ $a!=0$ , $x = \frac{- b \pm \sqrt{b^{2} - 4 a c}}{2 a}$ $x=(-b+-sqrt(b^2-4ac))/(2a)$ .

Here, $x = t, a = 5, b = - 22, c = 15$ $x=t,a=5,b=-22,c=15$ . Input:

$t = \frac{- (- 22) \pm \sqrt{{(- 22)}^{2} - 4 (5) (15)}}{2 \cdot 5}$ $t=(-(-22)+-sqrt((-22)^2-4(5)(15)))/(2*5)$

$t = \frac{22 \pm \sqrt{484 - 300}}{10}$ $t=(22+-sqrt(484-300))/10$

$t = \frac{22 + \sqrt{184}}{10}, \frac{22 - \sqrt{184}}{10}$ $t=(22+sqrt(184))/10, (22-sqrt(184))/10$

$t = \frac{22 + 2 \sqrt{46}}{10}, \frac{22 - 2 \sqrt{46}}{10}$ $t=(22+2sqrt(46))/10, (22-2sqrt(46))/10$

$t = \frac{11 + \sqrt{46}}{5}, \frac{11 - \sqrt{46}}{5}$ $t=(11+sqrt(46))/5, (11-sqrt(46))/5$

$t = 3.6 s, 0.8 s$ $t=3.6s,0.8s$

You get two answers because the ball reaches its topmost point, and then comes down again. This point can be found, it is the $x$ $x$ coordinate of the parabola of the function of $h (t)$ $h(t)$ :

graph{22x-5x^2 [-8.62, 11.38, 12.52, 22.52]}

As you can see, the topmost point of the stone will be at $2.2 s$ $2.2s$ , the average of our two answers (as it should).

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Question #5873d

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