A particle is released from a height $S$ above the surface of the earth. At a certain height, its kinetic ener

Question

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

Accepted Answer

1. **Conservation of Mechanical Energy:** The total mechanical energy ($E$) of the particle is conserved throughout the motion. At the release point (height $S$), the velocity is zero.
   $$ E = K_i + U_i = 0 + mgS = mgS $$
   (Assuming potential energy $U=0$ at the surface).
2. **Condition at Specific Height:** Let the particle be at height $h$ and have speed $v$. The problem states that Kinetic Energy ($K$) is three times Potential Energy ($U$).
   $$ K = 3U $$
3. **Calculate Height ($h$):**
   $$ E = K + U = 3U + U = 4U $$
   Substituting the expressions:
   $$ mgS = 4(mgh) \implies h = \frac{S}{4} $$
4. **Calculate Speed ($v$):**
   The Potential Energy at this height is $U = mg\left(\frac{S}{4}\right)$.
   The Kinetic Energy is $K = 3U = \frac{3mgS}{4}$.
   Using the formula $K = \frac{1}{2}mv^2$:
   $$ \frac{1}{2}mv^2 = \frac{3mgS}{4} $$
   $$ v^2 = \frac{3gS}{2} \implies v = \sqrt{\frac{3gS}{2}} $$

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