Calculating the Intersection of Two Stones Dropped from a Cliff

[John O' Meara]

A stone is droped from the top of a cliff of height 256 feet, and, at the same instant, another stone is projected vertically upwards from the ground with a speed of 96 ft/s. Find where they will meet?
I have the following:
v1=u1+at, v2=u2+at, s1=u1t+1/2at^2, s2=u2+1/2at^2,
I also have: v1^2=u1^2+2as1 and V2^2=u2^2+2as2
The question is what do I do next, please?

 

[Hootenanny]

You have the inital velocity and acceleration of both. You also know that the distance the stone thrown upwards from the ground has traveled when they meet is $256-s$ where $s$ is the distance traveled by the dropped stone.

 

[kyle8921]

To find where the two stones will meet, you can set the equations for their vertical positions (s1 and s2) equal to each other and solve for the time (t) at which they intersect. This can be done by substituting the equations for their vertical velocities (v1 and v2) and initial positions (u1 and u2) into the equations for displacement (s1 and s2).

Once you have solved for t, you can then plug that value back into the equation for s1 (or s2) to find the position at which the stones will meet. Keep in mind that you may need to use the quadratic formula to solve for t, as there may be multiple solutions.

Alternatively, you could also graph the equations for s1 and s2 and find the point at which they intersect on the graph. This would give you the position where the stones will meet.

It's also important to note that the equations you have listed may not be sufficient for this problem, as they do not take into account the acceleration due to gravity (g). You may need to modify them to include this acceleration in order to accurately calculate the intersection point of the two stones.