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## 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, v

_{iy}=20m/s, y=+5m(since it is an upward displacement), a=-9.81m/s²

(a) We use v²

_{fy}=v²

_{iy}+2ay to find

v²

_{fy}=(20m/s)²+2(-9.81m/s²)(5m) = 302m²/s²

v

_{fy}=±[tex]\sqrt{302m²/s²}[/tex]=-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=(v

_{fy}-v

_{iy})/t to find

t=[tex]\frac{(-17.4-20)m/s}{-9.81m/s²}[/tex]=3.8s

Notice that we retain the minus sign on v

_{fy}.

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 v

_{fy}=-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).

THANKS IN ADVANCE.