Given that:


u=Initial Velocity

v=Final Velocity



g=Gravitational Field Strength (modeled as 10m/s^2)

Assume lack of air resistance

A particle, 35 metres off of the ground, is projected upwards, and takes 4 seconds to fall to the ground.

The acceleration should therefore be -10, the total net displacement during that time should be -35, and the time is obviously 4 seconds.

Since s=vt - 1/2 at^2

-35=4v - 1/2*-10*4^2




However, according to my teacher, s is positive. I cannot comprehend this, since a is stated to be negative, and it is accelerating towards the ground, and s is a vector in the same direction as the acceleration.

His answer was something along the lines of s being the 'starting value'. Can anyone explain this to me?

EDIT: I put 10 instead of -10 for a


closed as off-topic by user99914, JMP, Mark, José Carlos Santos, Widawensen Jun 13 '18 at 11:56

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Your teacher used $a=g=10 m/s^2$, so your teacher used down is positive, and with the displacement being down as well, the displacement is therefore positive as well.

In your calculaton, you used $a=-10$, so you used down being negative, and therefore got a negative displacement ($-35$) as well.

In the end, you're both right; you just differed in what direction you considered positive.

  • $\begingroup$ No, my teacher used -10 for acceleration. $\endgroup$ – Piomicron Jun 12 '18 at 15:27
  • $\begingroup$ @Piomicron Hmmm... but you also said $g =10$, rather than $g=-10$ ... $\endgroup$ – Bram28 Jun 12 '18 at 15:34
  • $\begingroup$ The strength of g = 10, but the acceleration is acting downwards $\endgroup$ – Piomicron Jun 12 '18 at 15:36
  • $\begingroup$ @Piomicron And your teacher did the same thing, i.e. said $a = -g=-10$? Because if your teacher just said $a=g$, that would explain the difference... $\endgroup$ – Bram28 Jun 12 '18 at 15:36

It is mainly matter of convention, indeed we can assume any origin in teh space and any direction as positive arranging the equations accordingly.

In this case the teacher has assumed the origin at the ground and positive the upward direction for $s$ and thus the acceleration $a=-g$ has been assumed negative since it is directed downward.

Notably the equation for the motion is

  • $s(t)=s_0+v_0t+\frac12at^2=35+v_0t-\frac12gt^2$

with the condition that $s(4)=0$ we have

  • $s(4)=35+4v_0-80=0 \implies$ initial velocity $v_0=11.25 m/s$

which is positive accordingly to the convention adopted.

  • $\begingroup$ No, my teacher used -10 for acceleration. $\endgroup$ – Piomicron Jun 12 '18 at 15:27
  • $\begingroup$ @Piomicron Ah ok sorry I didn't get that. In this case your teacher assumed positive the upward direction for $s$ and the dislacement obtained is negative. $\endgroup$ – gimusi Jun 12 '18 at 15:29
  • $\begingroup$ If positive is the upward direction, and it's the downward direction with negative displacement, how was my answer incorrect? $\endgroup$ – Piomicron Jun 12 '18 at 15:33
  • $\begingroup$ @Piomicron $v_0$ is the initial velocity which is positive since the object was projected upward. If we choose $t=1$ we obtain $v_0=-25 m/s$ and in this case the object would be projected downward. $\endgroup$ – gimusi Jun 12 '18 at 15:37

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