A stone is dropped from a height h, while simultaneously another stone is thrown up from the ground and reaches a height of 4h. The question is, at what time do the two stones cross each other?



To solve this problem, let's first analyze the motion of each stone separately.

Stone A (Dropped Stone):
- Initially, Stone A is at a height h.
- It falls freely under the influence of gravity.
- Its motion can be described by the equation: h = (1/2) * g * t^2, where g is the acceleration due to gravity and t is the time.

Stone B (Thrown Stone):
- Initially, Stone B is at the ground level.
- It is thrown upwards with an initial velocity.
- Its motion can be described by the equation: h = v0 * t - (1/2) * g * t^2, where v0 is the initial velocity.



To find the time when the stones cross each other, we need to equate the heights of the two stones at a given time, t.

Setting h for Stone A equal to h for Stone B:
(1/2) * g * t^2 = v0 * t - (1/2) * g * t^2

Simplifying the equation:
g * t^2 = 2 * v0 * t

Now, let's solve for t:

g * t^2 - 2 * v0 * t = 0

Factoring out t:
t * (g * t - 2 * v0) = 0

From this equation, we have two possible solutions:
1. t = 0 (corresponding to the initial moment when the stones are released)
2. g * t - 2 * v0 = 0

Solving the second equation for t:
g * t = 2 * v0
t = (2 * v0) / g

Therefore, the stones cross each other at a time t = (2 * v0) / g.

Note: The value of v0 (initial velocity) and g (acceleration due to gravity) should be known to calculate the exact time when the stones cross each other.

The answer will be √h/8g