How to get an equation from a "ball is thrown" word problem?
Physics says two things. First, that the acceleration from gravity is constant, and second, that the acceleration is $-9.81 m/s^2$. You have accepted the first, which results in $y=ax^2+bx+c$ (assuming $x$ is time) but rejected the second. That is fine, you have enough data to evaluate all the constants, as you have three points on the parabola and need three constants, $a,b,c$. As you say, $c=60$ is immediate.
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Hamze
Updated on June 09, 2020Comments
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Hamze over 3 years
"A ball is thrown upward from the roof of a building $60$m tall. The ball reaches a height of $80$m above the ground after $2$ s and hits the ground $6$ s after being thrown." How do I determine the equation without using physics(constants related to physics) ideas?
It should be in the form $y= ax^2+ bx + c$, right? I got $60$ for the $y$-intercept so $y= ax^2+ bx + 60$. The ball had the height of $80$ meters at $2$ seconds. It dropped to the ground in $4$ seconds. So $\frac{80}{4}=b\implies b=20$? I found $a$ and got $-5x^2 + 20x + 60$ but do my steps show I understand the problem?
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Hamze about 7 yearsWe know the equation is y=ax^2+ bx +60. I'd put in the point (2,80) to get 80=a(2)^2 + b(2) +60. Put in the point (6,0) to get 0=a(6)^2 + b(6) +60. Now I have 2 equations with 2 unknowns. I used all 3 points to find all 3 constants by solving for a, and then b. Does this work?
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Ross Millikan about 7 yearsYes, that works fine. It turns out $b=20$, but your logic in the question does not justify that.
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Hamze about 7 yearsI don't understand why my method works, actually. Did I find the point of intersection? Could I have used any other 2 points?
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Ross Millikan about 7 yearsYou were lucky. It turns out $(2,80)$ is the peak, but that is not given. You can write it as $y=-5(x-2)^2+80$. If the ground were at the launch altitude the equation would be $y=-5(x-2)^2$ and it would have hit the ground at $4$ seconds. You would have found $b=20/2=10$ because the peak was $20$ and it hit the ground $2$ seconds later.
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Hamze about 7 yearsNo, I mean, the second method. The one where I put in (2,80) and (0,6). I don't understand why that works. Would you tell me?
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Ross Millikan about 7 yearsYou have three points on the parabola, if you add in $(0,60)$. That gives you three equations in the three unknowns $a,b,c$ You are just solving the three simultaneous equations to find them. As you noted in the question, the fact that you have a value at $x=0$ makes finding $c$ immediate.