# Vertically Thrown Stones

1. Oct 5, 2013

### postfan

1. The problem statement, all variables and given/known data
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)

2. Relevant equations

3. 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?

2. Oct 5, 2013

### haruspex

Did you try substituting those times back into the equations to see if they both produce 1.2m? (I did, and they do )

3. Oct 5, 2013

### postfan

So am I right?

4. Oct 5, 2013

yes.

5. Oct 6, 2013

### jee_van

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.

6. Oct 6, 2013