Crossing Point Distance Calculation

[zn23]

Homework Statement

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.

Homework Equations

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 don't know what to do next add the displacements or what ?

 

[Clairefucious]

Homework Statement

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.

Homework Equations

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 don't know what to do next add the displacements or what ?

The equation v = x/t is for constant velocities only, ie a = 0. So using it in this question is not going to help. You are looking for the point where the stones cross - at this point they will have the same displacement from the ground and will have traveled for the same length of time.

Use this information in your two relevant equations to find s = y up = (4 - y down) where s is the displacement from the ground at time t.

 

[zn23]

what wud u suggest another way to find time wud be ?