What Determines the Meeting Point of Two Falling Stones?

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Homework Help Overview

The problem involves two stones: one dropped from a height of 100 meters and another thrown downwards after a delay of 2 seconds. The stones are said to meet at a height of 21.6 meters above the ground. Participants are tasked with determining the time at which they meet, the initial speed of the second stone, and their speeds just before hitting the ground.

Discussion Character

  • Mixed

Approaches and Questions Raised

  • Participants discuss the equations of motion applied to both stones, with attempts to derive the time and initial speed based on their meeting height. Some participants express uncertainty about the calculations and question the values obtained, particularly regarding the time and height differences.

Discussion Status

There are varying calculations and interpretations of the problem. Some participants have provided alternative values for time and initial speed, suggesting a need for further verification of the original poster's computations. The discussion reflects a collaborative effort to clarify the problem setup and calculations without reaching a definitive conclusion.

Contextual Notes

Participants note discrepancies in the height calculations and the assumptions regarding the meeting point of the stones, indicating potential misunderstandings or miscalculations in the original problem statement.

Ab17
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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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