Stroboscopic photography physics

Because a small ball is released every once in a while, this photo is equivalent to taking a stroboscopic photo with an interval of 0. 1 second after only one small ball is released.

Because it is uniformly accelerated linear motion, the difference between adjacent displacements with equal spacing is constant, that is, BC-AB=CD-BC, so CD=25cm.

Proved as follows:

At a certain moment, the speed is V, after T, the speed is v+aT, and after T, the speed is v+2aT.

Therefore, the displacement difference between adjacent equal time intervals = [(v+2at) 2-(v+at) 2]/2a-[(v+at) 2-v2]/2a = at2.

So this is a fixed value.

According to this conclusion, the displacement difference =0.05m and T=0. 1s are introduced, and a = 5m/2 is obtained.

If the time to A is t, the time to B is t+0. 1, 0.5at2+0.15 = 0.5a (t+0.1) 2.

T = 0.25s seconds

And because T=0. 1s, there are two balls on it.

As for the scale problem you mentioned, it is because it is the law of uniform acceleration linear motion with zero initial velocity. You can find that 0. 1 is not divisible by 0.25, so if this "stroboscopic photo" is complete, it is not zero initial velocity, so it does not satisfy that proportional relationship.

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