Tracking a Falling Ball: Investigating Radius Over Time

AI Thread Summary
The discussion revolves around the behavior of a falling ball's radius as observed from a distance, specifically addressing how the radius appears to change over time. The equation of free fall, s = 1/2gt^2, is referenced to explain the relationship between time and the ball's position. Participants clarify that the radius does not physically decrease but rather that the visible angle subtended by the ball changes as it falls. The proposed formula, r(t) = r/(1/2gt^2 + x), is confirmed as a valid approximation for small radii when viewed from a significant distance. Overall, the conversation emphasizes understanding the visual perception of the ball's radius during free fall.
Numeriprimi
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Homework Statement


We have got a ball of radius r, which is falling from the roof of the house. How is the radius r with respect to time? We are looking at the ball directly from above and ball is at the beginning of fall x from our eyes. Neglect air resistance.

Homework Equations


I'm not sure what to write here. So I think the nub is equation of free fall: s= 1/2gt^2 radius decreases quadratically

The Attempt at a Solution


So, I know equation: s= 1/2gt^2
I think the radius decreases quadratically, because t^2.
I must attributed the x (s from eyes before free fall)
And my equation: r(t)= r/(1/2gt^2 + x)

It is right? Thanks for help.

PS: Sorry for bad English, but I don't know better yet.
 
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If r is the radius of the ball, I don't see any reason why it should decrease. Do you mean the visible angle for the observer?
If the initial distance of the ball is large compared to the radius of the ball, your r(t) is an approximation for that angle.

Hmm... the original problem statement would be interesting - with translated words maybe ;).
 
No, no... How changing the radius of sight of the observer, when ball is falling free fall.

Hmmm... You don't interesting what I think? :-( Describe better?
 
Hi Numeriprimi! :smile:
Numeriprimi said:
So, I know equation: s= 1/2gt^2
I think the radius decreases quadratically, because t^2.
I must attributed the x (s from eyes before free fall)
And my equation: r(t)= r/(1/2gt^2 + x)
mfb said:
… Do you mean the visible angle for the observer?
If the initial distance of the ball is large compared to the radius of the ball, your r(t) is an approximation for that angle.

i agree with mfb :smile: …

you seem to mean (half) the angle subtended at the eye, and your formula for r(t) confirms that (correctly, for small r) :wink:
 
So... Is my formula well?
 
your formula is good :smile:
 
Thread 'Variable mass system : water sprayed into a moving container'

Thread 'Variable mass system : water sprayed into a moving container'

Starting with the mass considerations #m(t)# is mass of water #M_{c}# mass of container and #M(t)# mass of total system $$M(t) = M_{C} + m(t)$$ $$\Rightarrow \frac{dM(t)}{dt} = \frac{dm(t)}{dt}$$ $$P_i = Mv + u \, dm$$ $$P_f = (M + dm)(v + dv)$$ $$\Delta P = M \, dv + (v - u) \, dm$$ $$F = \frac{dP}{dt} = M \frac{dv}{dt} + (v - u) \frac{dm}{dt}$$ $$F = u \frac{dm}{dt} = \rho A u^2$$ from conservation of momentum , the cannon recoils with the same force which it applies. $$\quad \frac{dm}{dt}...
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