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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?

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  • $\begingroup$ "Maybe that's an error?" Not in itself. But you have to be consistent. That means positions below the starting point (or wherever you have position zero) are negative, and downward accelrations must be negative always, while positions above zero, and upwards velocities and accelrations are positive. If you mix this up, you do get mistakes. $\endgroup$ – Arthur Sep 29 '17 at 21:45
  • $\begingroup$ if all the velocities and accelerations are in the same direction use the same sign. if they point in opposite direction use opposite signs. the sign here is really a convention used to indicate a direction. $\endgroup$ – shai horowitz Sep 29 '17 at 21:45
  • $\begingroup$ So, shouldn't I be able to use the s(t) equation to get 9.9s? I get the correct answer of 9.9 ONLY when I use the vfinal squared together with the vfinal=vinitial +2ad. $\endgroup$ – user163862 Sep 29 '17 at 21:53
  • $\begingroup$ @user163862 yes, it is $-2000 = 0 -44t + 1/2(-32)t^2$ $\endgroup$ – Nick Pavlov Sep 29 '17 at 21:55
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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 $s_0 = 0$.

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  • $\begingroup$ This is HUGE for me. My life just got simpler!!!! $\endgroup$ – user163862 Sep 30 '17 at 18:54

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