A stone thrown upwards with speed `u` attains maximum height `h`. Ahother stone thrown upwards from the same point with speed `2u` attains maximum height `H`. What is the relation between `h` and `H`?

Options

A. ` 2h= H`
B. ` 3h =H`
C. `4h=H`
D. `5h=H`

Explanation

To find the relationship between the maximum heights \( h \) and \( H \) attained by two stones thrown upwards with different initial speeds, we can use the equations of motion.

### Step-by-Step Solution:

1. **Identify the scenario**:
- A stone is thrown upwards with an initial speed \( u \) and reaches a maximum height \( h \).
- Another stone is thrown upwards with an initial speed \( 2u \) and reaches a maximum height \( H \).

2. **Use the equation of motion**:
The relevant equation of motion we will use is:
\[ v^2 = u^2 + 2as \]
where:
- \( v \) = final velocity (0 at maximum height),
- \( u \) = initial velocity,
- \( a \) = acceleration (which is \(-g\) due to gravity),
- \( s \) = distance traveled (height reached).

3. **For the first stone**:
- Initial speed \( u \), final speed \( v = 0 \), and height \( s = h \):
\[ 0 = u^2 - 2gh \]
Rearranging gives:
\[ u^2 = 2gh \quad \text{(Equation 1)} \]

4. **For the second stone**:
- Initial speed \( 2u \), final speed \( v = 0 \), and height \( s = H \):
\[ 0 = (2u)^2 - 2gH \]
Rearranging gives:
\[ 4u^2 = 2gH \quad \text{(Equation 2)} \]

5. **Relate the two equations**:
- From Equation 1, we have:
\[ u^2 = 2gh \]
- From Equation 2, we can express \( H \):
\[ H = \frac{4u^2}{2g} = \frac{2u^2}{g} \]

6. **Substituting Equation 1 into the expression for \( H \)**:
- Substitute \( u^2 = 2gh \) into \( H \):
\[ H = \frac{2(2gh)}{g} = 4h \]

7. **Conclusion**:
The relationship between the maximum heights \( h \) and \( H \) is:
\[ H = 4h \]

### Final Answer:
The relation between \( h \) and \( H \) is:
\[ H = 4h \]