# Uniformaly Accelerated Motion

## Main Question or Discussion Point

A stone is thrown straight upward with a speed of 20m/s. It is caught on its way down at a point 5m above where it was thrown. Find (a) How fast was it going when it was caught? (b) How long did the trip take?

Let us take up as positive. Then, for the trip that lasts from the instant after throwing to the instant before catching, viy=20m/s, y=+5m(since it is an upward displacement), a=-9.81m/s²
(a) We use v²fy=v²iy+2ay to find
fy=(20m/s)²+2(-9.81m/s²)(5m) = 302m²/s²
vfy=±$$\sqrt{302m²/s²}$$=-17m/s
We take the negative sign because the stone is moving downward, in the negative direction, at the final instant.

(b) We use a=(vfy-viy)/t to find
t=$$\frac{(-17.4-20)m/s}{-9.81m/s²}$$=3.8s
Notice that we retain the minus sign on vfy.

This question which are solved by my text book and I am having problem with them.
In part(b) for finding time(t) we take vfy=-17.4. Why it is negative? As we know that velocity can't be negative, if there is a negative sign so it just shows its direction, so we shouldn't take negative sign for finding time.
And also we are calculating time for whole trip so in part(b) we took, g=-9.81m/s², why this is negative too?
But if we draw figure then we see that when stone goes up then g will be negative while it comes down in its way then value of g will be taken as positive, right? If it is so, so how they took g=-9.81m/s²(which is shows stone goes upward while we are going to calculate time for whole trip in which for half way we take -g, when stone goes upward and in another half way we take +g, when stone comes down).

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Nabeshin
This question which are solved by my text book and I am having problem with them.
In part(b) for finding time(t) we take vfy=-17.4. Why it is negative? As we know that velocity can't be negative, if there is a negative sign so it just shows its direction, so we shouldn't take negative sign for finding time.
And also we are calculating time for whole trip so in part(b) we took, g=-9.81m/s², why this is negative too?
But if we draw figure then we see that when stone goes up then g will be negative while it comes down in its way then value of g will be taken as positive, right? If it is so, so how they took g=-9.81m/s²(which is shows stone goes upward while we are going to calculate time for whole trip in which for half way we take -g, when stone goes upward and in another half way we take +g, when stone comes down).
So the issue here is in choosing coordinate axes. The solutions seem to choose the up direction as positive and the down direction as negative. These are coordinate axes that exist independently of the stone or its motion. Therefore, since acceleration due to gravity is always down, its sign is always negative. The final velocity has negative sign since, as you point out, it is moving downwards at that point. The negative sign on this is very important! If you omit it, you are finding a different time, that is, the time at which the stone has slowed to 17.4m/s upwards (on the first half of the flight!).

Therefore, since acceleration due to gravity is always down, its sign is always negative.
No. If work done against gravity(upward) then we will take g as positive. But if work done along gravity(downward) then we will take g as negative. Don't you think that I am right?

The negative sign on this is very important! If you omit it, you are finding a different time, that is, the time at which the stone has slowed to 17.4m/s upwards (on the first half of the flight!).
If stone is thrown upward then its velocity is reduced to 0m/s not to 17.4m/s.

And I know that for downward acceleration due to gravity is +ve and for upward acceleration due to gravity is -ve.
My problem is stone goes up and then comes back down. When it goes up it has -g and when it comes down it has +g.
But when we are calculating time for whole trip so what sign should we taken with g?