The question asks how long it takes for a stone to fall from a height $4h$ if it takes $1s$ to fall from height $h$. Let's solve this step by step:

1. From the free-fall equation: The distance $h$ fallen by an object in time $t$ under gravity is given by: $h = \frac{1}{2} g t^2$ where $g$ is the acceleration due to gravity.

2. Find $h$ in terms of $g$: Given $t = 1s$ for height $h$, $h = \frac{1}{2} g (1)^2 = \frac{1}{2} g$

3. For height $4h$: Substituting $h = \frac{1}{2}g$ into $4h$, $4h = 4 \cdot \frac{1}{2} g = 2g$

   Using the free-fall equation for height $4h$, $2g = \frac{1}{2} g T^2$ where $T$ is the time to fall from $4h$.

4. Solve for $T$: $2g = \frac{1}{2} g T^2$ $T^2 = 4$ $T = 2 \, \text{seconds}$

Final Answer: B. 2s

Let me know if you'd like further clarification or additional examples!

5 Related Questions:

1. How does the time change if the height increases to $9h$?
2. What is the relationship between time and height in free fall?
3. How does the acceleration due to gravity influence the fall time?
4. What would be the fall time if $g = 9.8 \, \text{m/s}^2$?
5. Can air resistance affect the calculation of free-fall time?

One Tip:

Always remember, the time to fall is proportional to the square root of the height in free-fall motion without air resistance!