I struggle with the signs in these projectile questions. From a cliff $2000$ ft high a projectile is launched downward. The initial speed is given as $44$ ft/s. Since the projectile is launched downward, I assume the the initial velocity would be $-44$ f/s. Maybe that's an error? So, I easily get the correct answer of 9.9 seconds by first calculating the final velocity using: $v_f^2=v_0^2 + 2ad$, and then using this $v_f=-360.5f/s$ in the equation of $v_f=v_i +at$. BUT, in this last equation all my velocities (initial and final) and "$a$" are negative. $t=9.9 $ s.
I can immediately tell that I'm not fully understanding when to use positive and negative values for the acceleration.
In the equation of $v_f^2$ I used a negative value of initial velocity (however, since it was getting squared, it wouldn't matter if I had chosen positive) AND a positive value of acceleration. Wouldn't the acceleration and velocity have the same sign?
Now to add considerably to the confusion, I decided to try the problem in a shorter manner using the equation: $s(t)=s_0 + v_it-1/2at^2$.
For $v_i $ I use -44 f/s since the projectile is moving downward not upward. The problem states the projectile was launched downward. For acceleration I use +32, but since there is a negative sign in the equation I get $2000ft=0-44t-1/2(+32)t^2$. This equation produces imaginary results.
Ugh! Double Ugh! Is there an easier way to evaluate the signs of velocity and acceleration?
I always tell students to never write $s(t)=s_0+v_it-at^2/2$. Always write it with a plus, and then the $a$ should take care of it by being positive or negative depending on its direction. All your quantities which are actually vectors (so that includes $s$, $v$ and $a$ in all these kinematics problems) should be thought of with reference to a chosen direction. If you want $+$ to mean "up", then in this problem they should all be negative, because your object was thrown downward ($v_0 = -44$), the acceleration due to gravity is also downward ($g = -32$), and $s = -2000$ at the end of the fall because it is 2000ft below the initial position which you have taken to be $s_0 = 0$.