Two stones are thrown simultaneously, one straight upward from the base of a cliff and the other straight downward from the top of the cliff. The stones are thrown with the same speed.
The height of the cliff is 4 m and the speed with which the stones are thrown is 7 m/s.
find the distance of the crossing point above the ground.
gotta use these using the equations y up = v1t + 1/2at^2, and y down = v1t + 1/2at^2
The Attempt at a Solution
well so far i went v= x / t and found t = x / v, which was 0.6 s i plugged it in both equations, but i dont know what to do next add the displacements or what ????