A body is throw up with a velocity 'u'. It reaches maximum height 'h'. If its velocity of projection is doubled the maximum height it reaches is _____

Options

A. 4h
B. h
C. 2h
D. 3h

Answer

To solve the problem, we need to understand how the maximum height reached by a projectile is related to its initial velocity.

### Step-by-Step Solution:

1. **Understand the relationship between height and initial velocity**:
The maximum height \( h \) reached by a body thrown upwards with an initial velocity \( u \) can be derived from the equations of motion. The formula for maximum height is given by:
\[ h = \frac{u^2}{2g} \]
where \( g \) is the acceleration due to gravity.

2. **Determine the new initial velocity**:
If the initial velocity is doubled, the new initial velocity becomes:
\[ u' = 2u \]

3. **Calculate the new maximum height**:
Using the formula for maximum height with the new initial velocity \( u' \):
\[ h' = \frac{(u')^2}{2g} = \frac{(2u)^2}{2g} \]

4. **Simplify the expression**:
Expanding the equation:
\[ h' = \frac{4u^2}{2g} = \frac{2u^2}{g} \]

5. **Relate the new height to the original height**:
From the original height \( h = \frac{u^2}{2g} \), we can express \( h' \) in terms of \( h \):
\[ h' = 4 \left(\frac{u^2}{2g}\right) = 4h \]

6. **Conclusion**:
Therefore, if the velocity of projection is doubled, the maximum height it reaches is:
\[ \text{Maximum height} = 4h \]

### Final Answer:
The maximum height it reaches is \( 4h \).

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