What Determines the Meeting Point of Two Falling Stones?

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The discussion revolves around a physics problem involving two stones dropped from a building. The first stone is dropped from a height of 100 meters, and the second stone is thrown downwards 2 seconds later, passing the first stone at a height of 21.6 meters above the ground. Participants calculated the time when the stones meet and the initial speed of the second stone, with one user suggesting an initial speed of 30.19 m/s and another proposing 29.4 m/s. There were discrepancies in the calculations regarding the heights and times, prompting a call for verification of the computations. The conversation highlights the importance of accuracy in solving kinematics problems.
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Homework Statement



A stone is dropped from rest from the roof of a 100-m high building; 2.00 s after that, a second stone is thrown straight down with initial speed v0, and the second stone passes the first stone 21.6 m above ground. (a) When are the stones at the same height? (b) What was the initial speed, v0, of the second stone, in m/s? (c) What are the speeds of the two stones, in m/s, just before they hit the ground?

h= 100m
h(meet)= 26.1m
V1i= 0m/s

Homework Equations


Xf=Xi +Vt +0.5at^2

The Attempt at a Solution


(a). Xf1= -4.9t^2
Xf2 = Vo(t-2) -4.9(t-2)^2
-73.1 = -4.9t^2

(b). -73.1 = Vo(1.86) -4.9(1.86)^2
Vo = 30.19 m/s

(c). Vf = Vi + at
Vf = 0 -9.8(3.86)
Vf = 37.83 m/s

Vf= 30.19 -9.8(1.86)
Vf= 48.42m/s

Am I right? Not sure if I am
 
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Ab17 said:

Homework Statement



A stone is dropped from rest from the roof of a 100-m high building; 2.00 s after that, a second stone is thrown straight down with initial speed v0, and the second stone passes the first stone 21.6 m above ground. (a) When are the stones at the same height? (b) What was the initial speed, v0, of the second stone, in m/s? (c) What are the speeds of the two stones, in m/s, just before they hit the ground?

h= 100m
h(meet)= 26.1m
V1i= 0m/s

Homework Equations


Xf=Xi +Vt +0.5at^2

The Attempt at a Solution


(a). Xf1= -4.9t^2
Xf2 = Vo(t-2) -4.9(t-2)^2
-73.1 = -4.9t^2
You don't give the time, although the next entry seems to have used the correct time.

(b). -73.1 = Vo(1.86) -4.9(1.86)^2
Vo = 30.19 m/s

(c). Vf = Vi + at
Vf = 0 -9.8(3.86)
Vf = 37.83 m/s

Vf= 30.19 -9.8(1.86)
Vf= 48.42m/s

Am I right? Not sure if I am
You calculated the speed of the both stones at 21.6 m above the ground.
 
Last edited:
I could be wrong, but I found slightly different values. For time, t = 4s, and initial speed V0 = 29,4. Are you sure about your results?
 
ramzerimar said:
I could be wrong, but I found slightly different values. For time, t = 4s, and initial speed V0 = 29,4. Are you sure about your results?
I think you're right.

I didn't even think to check the distances OP came up with.

100 - 21.6 = 78.4, not 73.1

If it's supposed to be 26.1,
then 100 - 26.1 = 73.9, not 73.1
 
Something isn't adding up here:
Ab17 said:
100-m high building

Ab17 said:
21.6 m above ground

Ab17 said:
h(meet)= 26.1m

Ab17 said:
-73.1
 
@Ab17 double check your computation in (a).
 
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