Rocks falling at the same time and their distance apart

[Glyper]

Homework Statement

Two rocks are falling from a roof. The second one begins its fall when the first one has already covered 30cm. Calculate the distance d between these two rocks after 10 seconds counted from the moment the second rock begins its fall.

Homework Equations

$s=s_{0}+v_{0}t+\frac{at^{2}}{2}$
$v(t) = v_0 + at$

The Attempt at a Solution

It turns out second rock will go as follows: $s_2=\frac{at^2}{2}={9,81*100}{2}\approx 490,5m$. If it does, then we have $0,3=\frac{9,81t^2}{2}$ so $t\approx0,25s$. Then, we know that $v(0,25)=0+9,81*0,25=2,4525m/s$. And finally $s_1=0,3m+2,45*10+\frac{9,81\cdot100}{2}=515,3m$ so they're 24,8 metres apart. Is that OK?

 

[Glyper]

Is something wrong here? I can't find any but something surely does not feel right.

 

[mishek]

What seems wrong?

 

[lendav_rott]

I got to the same result.
Basically you compare distances.
s1 = Δs + g(0.247+10)²/2
s2 = 0 + g10²/2

s1 - s2 = Δs + g/2(10.247²-10²) = 0.3 + 24.529 ~ 24.8
Δs is the freefall during the 1st 0.247 or 0.25 seconds - the 30cm given.

I am fairly certain you have done it correctly.

 

[Glyper]

Oh... That's good. Thank you :) Something felt wrong as I'm simply not too confident with my physics skills ;)

 

[mishek]

I got to the same result.
Basically you compare distances.
s1 = Δs + g(0.247+10)²/2
s2 = 0 + g10²/2

s1 - s2 = Δs + g/2(10.247²-10²) = 0.3 + 24.529 ~ 24.8
Δs is the freefall during the 1st 0.247 or 0.25 seconds - the 30cm given.

I am fairly certain you have done it correctly.

I'm having trouble understanding the first line - s1 = Δs + g(0.247+10)²/2. Isn't Δs already taken into account with these 0.247 sec?

 

[lendav_rott]

Can't find a comprehendable explanation - my lingual skills fail me :(

The way I understand it is that if I plug 10s into both of these equations, they both cancel out and the constant distance between them is always 30cm - but that makes no sense, because the 1st stone already has a base velocity of 2.something metres per second and the only way to express it in the equation is to use the time on top of the 10s to show that s1 doesn't start at the same speed as s2.

 

[mishek]

Please take into consideration following:

10.25^2*9.81*0.5-10^2*9.81*0.5=24.83.

10.247^2*9.81*0.5-10^2*9.81*0.5=24.52.

 

[voko]

Since the acceleration is the same in both cases, the only difference in distances will be due to the initial difference (30 cm) and the initial velocity of the first rock (2.45 m/s according to your calculation). Thus diff = 30 cm + 2.45 m/s * 10 s = 24.8 m.

 

[HallsofIvy]

Here's how I would have done the problem: Since the first rock is dropped its initial speed is 0 and so the distance dropped is $d_1= (g/2)t^2$. It will have dropped 30 m when $(g/2)t^2= 30$ so $t= \sqrt{60/g}$.

The second rock is dropped at that time also with initial speed 0 and so $d_2= (g/2)(t- \sqrt{60/g})^2= (g/2)t^2- g\sqrt{60/g}t+ 30$.

Subtracting, the distance between the two rocks, after t seconds, is $g\sqrt{60/g}t- 30$.