Question

A particle is released from height S from the surface of the Earth. At a 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. \(\frac{S}{4},\sqrt{\frac{3gS}{2}}\)
• B. \(\frac{S}{4},{\frac{3gS}{2}}\)
• C. \(\frac{S}{4},{\frac{\sqrt{3gS}}{2}}\)
• D. \(\frac{S}{2},{\frac{\sqrt{3gS}}{2}}\)

Answer 1

The Correct Option is A

To solve this problem, we need to determine the height from which the kinetic energy (KE) of the particle is three times its potential energy (PE) and find the speed at that instant.

Let's denote:

• \( h \) as the height from the surface of the Earth at which this condition is met,
• \( S \) as the initial height from which the particle is released,
• \( g \) as the acceleration due to gravity,
• \( m \) as the mass of the particle.

The gravitational potential energy at height \( h \) is given by:

\(PE = mgh\)

The kinetic energy of the particle is given by:

\(KE = \frac{1}{2}mv^2\)

According to the problem, the kinetic energy is three times the potential energy:

\(\frac{1}{2}mv^2 = 3mgh\)

From this equation, we can solve for the velocity:

\(v^2 = 6gh\)

\(v = \sqrt{6gh}\)

Since the particle was released from rest, the total energy at height \( S \) is purely potential:

\(PE_{\text{initial}} = mgS\)

At height \( h \), using the conservation of mechanical energy, the potential energy plus kinetic energy should equal the initial potential energy:

\(mgh + \frac{1}{2}mv^2 = mgS\)

Substituting the expression for \( v^2 \):

\(mgh + 3mgh = mgS\)

\(4mgh = mgS\)

\(h = \frac{S}{4}\)

Therefore, the height is \( \frac{S}{4} \).

Substituting \( h = \frac{S}{4} \) into the expression for \( v \):

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

Thus, the speed of the particle at that height is \( \sqrt{\frac{3gS}{2}} \).

Hence, the correct answer is:

\(\frac{S}{4},\sqrt{\frac{3gS}{2}}\)