Calculating Vertical Stone Throw Intersection Time and Altitude

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Homework Help Overview

The discussion revolves around calculating the intersection time and altitude of two stones thrown vertically from the ground, considering gravitational acceleration. The first stone is thrown with an initial speed, followed by a second stone thrown after a delay.

Discussion Character

  • Exploratory, Problem interpretation, Assumption checking

Approaches and Questions Raised

  • Participants discuss the equations of motion for both stones and the conditions under which they meet at the same altitude. There are attempts to verify calculations by substituting values back into the equations.

Discussion Status

Some participants have provided feedback on the calculations, confirming that the values substituted into the equations yield consistent results. Others have posed questions about the correctness of the approach and the need for further clarification.

Contextual Notes

There are multiple instances of similar problems being presented, indicating a potential confusion or overlap in understanding the problem setup. Participants are encouraged to clarify their queries and avoid redundancy in thread topics.

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Homework Statement


A stone is thrown up vertically from the ground (the gravitational acceleration is g=10 m/s2). After a time Δt= 1 s, a second stone is thrown up vertically. The first stone has an initial speed v1= 7.0 m/s, and the second stone v2= 7.0 m/s.

(a) At what time t after the first stone is thrown will the two stones be at the same altitude h above ground? (in seconds)

(b) At what altitude h above ground will the two stones meet? (in meters)

Homework Equations


The Attempt at a Solution


I created two equations : h=-.5gt^2+vt and h=-.5g(t-1)^2+v(t-1).
Solved for t and got 1.2, then substituted t=1.2 in the equations and got h=1.2. Is that right?
 
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Did you try substituting those times back into the equations to see if they both produce 1.2m? (I did, and they do :smile:)
 
So am I right?
 
postfan said:
So am I right?
yes.
 
A stone is thrown up vertically from the ground (the gravitational acceleration is g=10 m/s2). After a time Δt= 1 s, a second stone is thrown up vertically. The first stone has an initial speed v1= 13.0 m/s, and the second stone v2= 18.0 m/s.

(a) At what time t after the first stone is thrown will the two stones be at the same altitude h above ground? (in seconds)
(b) At what altitude h above ground will the two stones meet? (in meters)

h=?

Friends, i would really appreciate if you help me out to solve this question.
 
jee_van said:
A stone is thrown up vertically from the ground (the gravitational acceleration is g=10 m/s2). After a time Δt= 1 s, a second stone is thrown up vertically. The first stone has an initial speed v1= 13.0 m/s, and the second stone v2= 18.0 m/s.

(a) At what time t after the first stone is thrown will the two stones be at the same altitude h above ground? (in seconds)
(b) At what altitude h above ground will the two stones meet? (in meters)

h=?

Friends, i would really appreciate if you help me out to solve this question.

hey, start a new/separate thread!
 

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