Calculating Vertical Stone Throw Intersection Time and Altitude

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The discussion revolves around calculating the intersection time and altitude of two vertically thrown stones under gravitational acceleration. The first stone is thrown with an initial speed of 7.0 m/s, while the second is thrown one second later with the same speed. The solution involves setting up equations for the height of each stone and solving for the time and altitude where they meet. The initial calculations suggest that the stones meet at approximately 1.2 seconds after the first stone is thrown and at an altitude of 1.2 meters. Participants confirm the correctness of these calculations and encourage others to start new threads for different questions.
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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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