Question

A body when projected at an angle \( \theta \) with the horizontal reaches a maximum height \( H \). The time of flight of the body will be ( \( g \) = acceleration due to gravity)

• A. \( \frac{1}{2} \sqrt{\frac{2H}{g}} \)
• B. \( \sqrt{\frac{g}{2H}} \)
• C. \( 2 \sqrt{\frac{2H}{g}} \)
• D. \( \sqrt{\frac{2H}{g}} \)

Hint:

Time of flight is exactly twice the time taken to reach the maximum height.

Answer 1

The Correct Option is C

Step 1: Formula for H
$H = \frac{u^2 \sin^2 \theta}{2g} \Rightarrow u \sin \theta = \sqrt{2gH}$
Step 2: Formula for Time of Flight
$T = \frac{2u \sin \theta}{g}$
Step 3: Substitution
$T = \frac{2 \sqrt{2gH}}{g} = 2 \sqrt{\frac{2gH}{g^2}} = 2 \sqrt{\frac{2H}{g}}$
Step 4: Conclusion
The time of flight is $2 \sqrt{2H/g}$.
Final Answer:(C)