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: Understanding the Question:
We need to express the time of flight in terms of the maximum height achieved by a projectile.
Step 2: Key Formula or Approach:
Max Height \( H = \frac{u^2 \sin^2 \theta}{2g} \).
Time of Flight \( T = \frac{2 u \sin \theta}{g} \).
Step 3: Detailed Explanation:
From the height formula:
\[ u \sin \theta = \sqrt{2gH} \] Substitute this into the time of flight formula:
\[ T = \frac{2 \sqrt{2gH}}{g} = 2 \sqrt{\frac{2gH}{g^2}} = 2 \sqrt{\frac{2H}{g}} \] Step 4: Final Answer:
The time of flight is \( 2 \sqrt{\frac{2H}{g}} \).