Question

A particle is released from height S above the surface of the earth. At certain height its kinetic energy is three times its potential energy. The height from the surface of the earth and the speed of the particle at that instant are respectively.

• A. \( \frac{S}{2} \), \( \sqrt{\frac{3gS}{2}} \)
• B. \( \frac{S}{2} \), \( \frac{3gS}{2} \)
• C. \( \frac{S}{4} \), \( \sqrt{\frac{3gS}{2}} \)
• D. \( \frac{S}{4} \), \( \frac{3gS}{2} \)

Hint:

Use the conservation of energy principle to relate the potential and kinetic energies at different heights. Remember that the potential energy is proportional to the height above the surface.

Answer 1

The Correct Option is C

This problem is solved by analyzing the given conditions alongside the equations of motion and energy.

1. \(S\) represents the initial release height. At this height, the particle's potential energy (PE) is at its maximum, and its kinetic energy (KE) is zero.
2. As the particle descends, its PE diminishes while its KE escalates. At an arbitrary height \(h\), the particle's kinetic energy is stipulated to be thrice its potential energy.

The formulas for potential and kinetic energy are as follows:

• Potential energy at height \(h\): \(PE = mgh\)
• Kinetic energy at height \(h\): \(KE = \frac{1}{2}mv^2\)

The problem states that \(KE = 3 \times PE\).

\(\frac{1}{2}mv^2 = 3 \times mgh\)
\(v^2 = 6gh\) (Equation 1)

By the principle of conservation of energy, the total mechanical energy at the start equals the total mechanical energy at height \(h\).

• Total energy at height \(S\): \(mgS\) (initial KE is 0)
• Total energy at height \(h\): \(mgh + \frac{1}{2}mv^2\)

Equating the initial and current energies:

\(mgS = mgh + \frac{1}{2}mv^2\)
\(mgS = mgh + 3mgh\) (since KE = 3PE)
\(mgS = 4mgh\)
\(gS = 4gh\)
\(h = \frac{S}{4}\) (Equation 2)

Substituting Equation 2 into Equation 1 yields the velocity:

\(v^2 = 6g \left(\frac{S}{4}\right)\)
\(v^2 = \frac{3gS}{2}\)
\(v = \sqrt{\frac{3gS}{2}}\)

Consequently, the particle is at a height of \(\frac{S}{4}\) from the surface, and its speed is \(\sqrt{\frac{3gS}{2}}\).