A stone is projected vertically upward from some height at $$2v$$. Some $$\Delta t$$ later another stone is projected vertically upward at $$v$$ from the same position. If they reach the initial position at the same time, find $$v$$.

PLACEHOLDER ANSWER - the answer is an expression. A stone thrown up at speed $$u$$ returns to its starting point after $$\frac{2u}{g}$$. So the first returns after $$\frac{4v}{g}$$ and the second after $$\Delta t + \frac{2v}{g}$$. Setting these equal gives $$\frac{2v}{g} = \Delta t$$, so $$v = \frac{g\,\Delta t}{2}$$.