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I see many examples of projectile motion equations using $-16$ as the $a$ coefficent. For example:

$$-16t^2+36t+50$$

Why is this?

SpecialBomb
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3 Answers3

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I don't know why this perfectly reasonable question has downvotes.

I also don't know your math background so I'll explain this under the assumption that you don't know calculus.

In Imperial units, acceleration due to gravity is approximately $-32 \text{ ft/s}^2$. So we can define a constant function $a(t)$ for the gravitational acceleration of an object at time $t$. It would be $a(t) = -32$.

In calculus there's an operation called integration. You can integrate the acceleration $a(t)$ to get the velocity $v(t)$. When we integrate our $a(t)$ from the previous paragraph, we get $v(t) = -32t + C$, where $C$ represents some arbitrary constant which can be determined from additional information that would be given in a specific problem.

Then we can integrate the velocity $v(t)$ to get the position $s(t)$. When we integrate our $v(t)$ from the previous paragraph, we get $s(t) = -16t^2 + Ct + B$, where $B$ is another arbitrary constant.

That's where the $-16$ comes from. Integrating the acceleration function, then integrating the velocity function.

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    I think people downvoted this because it is more of a science/physics question than a math question, so most people would probably see this as question not suitable for the site. I think it's a good question, though. – Noble Mushtak Apr 20 '16 at 13:32
  • Makes sense. I guess I've just taught precalculus so much that I automatically link projectile motion to math instead of physics. –  Apr 20 '16 at 13:33
  • Other reasons: There is no equation, there is no explanation of 'a coefficient'. – gammatester Apr 20 '16 at 13:37
  • Eh. The implied equation should be clear to anyone who remembers projectile motion, and since OP gave a quadratic expression it can be reasonably assumed that $a$ is the coefficient on the squared term. Also OP did say "using $-16$ as the $a$ coefficient" so I'm not sure how there could be enough confusion for a downvote. Maybe people would've been happier with, e.g., "using $a = -16$ where $a$ is the coefficient on $t^2.$" –  Apr 20 '16 at 13:44
  • And if that's the case then it just reinforces my feeling that sometimes we expect too much of people who ask questions here. A lot of them are students just learning these things for the first time, so it's unreasonable to expect them to be able to communicate these things in the same way that professionals or experts do. –  Apr 20 '16 at 13:45
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It's because the acceleration of an object because of gravity on Earth in imperial units is approximately $-32 \frac{\text{ft}}{\text{s}^2}$ and the coefficient of $t^2$ is $\frac 1 2 g=-16$.

If you are using metric, you will also see $-4.9$ a lot because in metric, $g \approx -9.8 \frac{\text{m}}{\text{s}^2}$, so the coefficient of $t^2$ is $-4.9$.

Noble Mushtak
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It's probably because in "english" units, the acceleration of gravity is (approximately) 32 ft/sec^2. When you integrate this twice, you get a factor of 1/2 along the way, resulting in 16.

John Hughes
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