Why do so many projectile motion equation examples use $-16$ as the $a$ coefficient? I see many examples of projectile motion equations using $-16$ as the $a$ coefficent. For example:
$$-16t^2+36t+50$$
Why is this?
 A: 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.
A: 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 $-4.9$.
A: 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. 
