Question

A stone is thrown at an angle \(\theta\) with the horizontal reaches a maximum height \(H\), then the time of flight of stone will be (\(g\) = acceleration due to gravity)

• A. \(\left(\frac{2H}{g}\right)^{\frac{1}{2}}\)
• B. \(\frac{2(2H\sin \theta )^{\frac{1}{2}}}{g}\)
• C. \(2\left(\frac{2H}{g}\right)^{\frac{1}{2}}\)
• D. \(\frac{(2H\sin \theta )^{\frac{1}{2}}}{g}\)

Hint:

Express the launch speed component in terms of the maximum height, then substitute into the time of flight formula.

Answer 1

The Correct Option is C

Step 1: Understanding the Concept
For a projectile with speed \(u\) at angle \(\theta\), the maximum height is \(H=\dfrac{u^{2}\sin^{2}\theta}{2g}\) and the time of flight is \(T=\dfrac{2u\sin\theta}{g}\).

Step 2: Eliminate u
From the height formula, \(u\sin\theta=\sqrt{2gH}\).

Step 3: Calculation
\(T=\dfrac{2\sqrt{2gH}}{g}=2\sqrt{\dfrac{2H}{g}}\)

Step 4: Compare the options
This equals \(2\left(\dfrac{2H}{g}\right)^{1/2}\), option (C).

Final Answer:
The time of flight is \(2\sqrt{2H/g}\). \[ \boxed{\text{(C) }2\left(\dfrac{2H}{g}\right)^{\frac{1}{2}}} \]