## Physical Significance of Numbers

Consider this case: A ball is allowed to fall freely with a constant acceleration of 10 m/s^2. What will its speed be after travelling a distance of 10 m?

We get that:
v[initial] = 0 m/s
d = -5 m (downward direction is taken as negative)
a = -10 m/s^2

To find v[final], we will use the formula d = (v[final]^2-v[initial]^2) / (2a)

Rearranging, we will get:

By plugging the numbers, we will get:

v[final]^2 = 100 m^2/s^2

By mathematics, there are 2 solutions to this problem, that is v[initial][1] = +10 m/s and v[initial][2] = -10 m/s. I can only think of the situation where the ball has a downward velocity. Is the other answer (the one with +10 m/s velocity) possible/does it has a physical reality? If so, can anyone explain the situation? Thanks a lot!

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 I think to obtain your first formula, the identity vf - vi = a*t is used. Next, you divide by a to get t, and plug that into d = a/2 * t^2. But strictly speaking, d, v, and a are vectors. So you are in fact not allowed to divide by a. You could work around this by saying you just talk about one component (the z-component) of each vector. But this implies that (vf-vi) has the same sign as a (i.e., minus). Of course, this is still true in the answer.
 Hi, u consider a stationary ball 'b' directly below the falling ball and calculate the velocity of 'b' in + with respect to free falling ball. Else, if free fall is not crucial, vertical take off of a rocket (containing the ball) with acceleration 10 m/s^2. v = u + at = final velocity of both rocket & ball.

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## Physical Significance of Numbers

 By mathematics, there are 2 solutions to this problem, that is v[initial][1] = +10 m/s and v[initial][2] = -10 m/s. I can only think of the situation where the ball has a downward velocity. Is the other answer (the one with +10 m/s velocity) possible/does it has a physical reality? If so, can anyone explain the situation? Thanks a lot!

The -10 m/s velocity occurs when you trace the evolution of the system going forward in time. Consider the trajectory of the ball before it had the 0 m/s velocity.

Hurkyl

 Aaaah..... I got it! Thanks a lot Hurkyl!