What Determines the Meeting Point of Two Falling Stones?

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SUMMARY

The discussion centers on the physics problem involving two stones dropped from a height of 100 meters. The first stone is dropped from rest, while the second stone is thrown downwards with an initial speed, v0, 2 seconds later. The stones meet at a height of 21.6 meters above the ground. The calculations reveal that the initial speed v0 of the second stone is approximately 30.19 m/s, and the speeds of both stones just before impact are 37.83 m/s and 48.42 m/s, respectively. However, discrepancies in height calculations indicate a need for careful verification of the values used in the equations.

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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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