A stone is dropped from a cliff; 2.18 s later another stone is thrown downward with an initial speed of 31 m/s. They reach the ground simultaneously. Find the height of the cliff.
vf^2 = vo^2 + 2ax
x = vot + 0.5at^2
vf = vo + at
The Attempt at a Solution
i've been working on this for the past two hours, and it's driving me nutsss.
i found the speed of the stone in 2.18 seconds by using the equation vf = vo + at. (t = 2.18 seconds; a = 9.81 m/s^2; vo = 0) i found the velocity to be 21.3858 m/s.
with that information, i found how far the stone traveled with the equation vf^2 = vo^2 + 2ax. (vf^2 = (21.3858)^2; vo^2 = 0; a = 9.81) the answer: 23.310522 m.
after that, i don't know what else to do with stone b. i've been playing around with the first equation, since the two stones hit the ground at the same time. is it wrong to assume that the final velocities of the two would be the same?
i would appreciate it very much if someone could throw me a bone on what to do with the second stone because i'm at a loss.