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 find the height from the surface of the earth at which the kinetic energy (KE) of the particle is three times its potential energy (PE). We will also determine the speed of the particle at that height.

Initially, the particle is released from a height \(S\) from the surface of the Earth with zero initial velocity. The total mechanical energy (TE) at the height \(S\) is given by:

\(PE_{\text{initial}} = m \cdot g \cdot S\) and \(KE_{\text{initial}} = 0\). 

Thus, \(TE_{\text{initial}} = m \cdot g \cdot S\).

At a certain height \(h\) from the surface, we have:

• \(PE = m \cdot g \cdot h\)
• \(KE = \frac{1}{2} \cdot m \cdot v^2\)

According to the problem, \(KE = 3 \cdot PE\). Therefore:

\(\frac{1}{2} \cdot m \cdot v^2 = 3 \cdot m \cdot g \cdot h\)

Simplifying, we get:

\(v^2 = 6 \cdot g \cdot h\)

Since the total mechanical energy is conserved, we have:

\(m \cdot g \cdot S = m \cdot g \cdot h + \frac{1}{2} \cdot m \cdot v^2\)

Substitute \(\frac{1}{2} \cdot m \cdot v^2 = 3 \cdot m \cdot g \cdot h\) into the above conservation equation:

\(m \cdot g \cdot S = m \cdot g \cdot h + 3 \cdot m \cdot g \cdot h\)

\(m \cdot g \cdot S = 4 \cdot m \cdot g \cdot h\)

Canceling the common terms and solving for \(h\), we get:

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

Now, substitute \(h = \frac{S}{4}\) back into \(v^2 = 6 \cdot g \cdot h\) to find the speed:

\(v^2 = 6 \cdot g \cdot \frac{S}{4}\)

\(v^2 = \frac{3 \cdot g \cdot S}{2}\)

\(v = \sqrt{\frac{3 \cdot g \cdot S}{2}}\)

Thus, the height from the surface of the Earth and the speed of the particle at that instant are respectively \(\frac{S}{4}\) and \(\sqrt{\frac{3gS}{2}}\), which matches the correct option:

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