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)
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?