Why do so many projectile motion equation examples use $-16$ as the $a$ coefficient?

Question

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

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

Why is this?

Answer 1

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.

Answer 2

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 $-4.9$.

Answer 3

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.