Question

A body is thrown from the surface of the Earth with velocity \( V \, \text{m/s} \). The maximum height above the Earth's surface up to which it will reach is
\textit{(R = radius of earth, \( g \) = acceleration due to gravity)}

• A. \( \frac{VR^2}{gR - V} \)
• B. \( \frac{V^2R}{2gR - V^2} \)
• C. \( \frac{2gR}{V^2(R - 1)} \)
• D. \( \frac{VR}{2gR - V} \)

Hint:

When a body is thrown upwards from Earth's surface, use energy conservation to relate the initial velocity and maximum height.

Answer 1

The Correct Option is B

Step 1: Formula for maximum height.
The maximum height \( h \) reached by the body is determined by the equation: \[ h = \frac{V^2R}{2gR - V^2} \] This formula comes from equating the initial kinetic energy to the work done by gravity.
Step 2: Conclusion.
Thus, the correct answer is (B) \( \frac{V^2R}{2gR - V^2} \).