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A particle $A$ is dropped from height $100\ m$ and another particle $B$ is projected vertically up with velocity $50\ m/s$ from the ground along the same time. Find out the position where two particle will meet? (take $g=10\ m/s^2$)

Book sol.

They took upward direction to be positive.

For particle $A$,

$y_0=+100\ m \\ u=0\ m/s \\ a=-10\ m/s^2$

For particle $B$,

$y_0=0\ m \\ u=+50\ m/s \\a=-10\ m/s^2$

Now my doubt is if positive is upwards then particle $B$ is accelerating along it's position vector (from origin, which is where our particle $B$ lies), then why $a=-10\ m/s^2$ not $a=+10\ m/s^2$.

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  • $\begingroup$ The projection of $B$ upwards does not give it an acceleration upwards (beyond the initial acceleration from $0$, which you can assume happens instantaneously), but a velocity upwards. At the moment of its release, it is moving the fastest upwards that it will ever move upwards and immediately begins slowing down, or accelerating in the negative direction, due to gravity. $\endgroup$
    – userManyNumbers
    Commented May 21, 2017 at 13:50
  • $\begingroup$ @user3949312 if so then why $A$'s is negative? $\endgroup$
    – mathlover
    Commented May 21, 2017 at 13:56
  • $\begingroup$ Acceleration is negative simply by the convention used in this question; it just defined downwards to be negative and upwards positive. Downwards and upwards don't need to be negative and positive respectively, as long as they are of opposite sign to each other. If you invert all the signs and do the calculations, you'll find that the two particles will still meet at the same place (though the spatial position will also have the sign reversed). $\endgroup$
    – userManyNumbers
    Commented May 21, 2017 at 14:01
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    $\begingroup$ (The sign communicates direction, not magnitude. If acceleration occurs in the downwards direction, it is given a negative sign. This doesn't mean the object is slowing down, but that it is accelerating in the downwards direction. It could be speeding up or slowing down, depending on the direction in which it was already moving. If it was already moving downwards ($-v$), the $-a$ downwards will add to the $-v$ over time and give an even more negative $-v$, and it will move downwards faster. If it was moving upwards ($+v$), adding the $-a$ over time will decrease its $+v$.) $\endgroup$
    – userManyNumbers
    Commented May 21, 2017 at 14:11

2 Answers 2

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The acceleration due to gravity is always downwards (which is why a = -10 m/s$^2$ for both particles).

The initial velocities may be either up, down, or zero (e.g. positive or negative or zero), but the Earth always pulls (accelerates) particles downward.

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  • $\begingroup$ is it merely a convention or can be explained, if later then please do. $\endgroup$
    – mathlover
    Commented May 21, 2017 at 13:53
  • $\begingroup$ The notation you used is that up is the positive direction (and down is negative). As far as the particles are concerned, they are both pulled downward by the Earth. This is why their acceleration due to the Earth's gravity is negative. Your statement that the 'acceleration is along the position vector' is incorrect. $\endgroup$
    – Bob
    Commented May 21, 2017 at 14:00
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The acceleration due to gravity is always directed towards the centre of Earth.If you go by its formal meaning it is the Gravitational acceleration, the acceleration caused by the gravitational attraction of Earth.

Earth Pulls each and every body in your case its downward.Added to this ,the upward projection of B imparts only velocity not acceleration to the body.It is finally the Earth which pulls down both the bodies and imparts a negative acceleration according to your sign convention.

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  • $\begingroup$ means $g$ will always have -ve sign, no matter where or object is going? $\endgroup$
    – mathlover
    Commented May 21, 2017 at 14:16
  • $\begingroup$ Yes you are right if only if you take g as -ve. $\endgroup$
    – DarkSideofPhy
    Commented May 25, 2017 at 16:50

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