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.
$\frac{S}{4}$, $\sqrt{\frac{3gS}{2}}$
Particle Release from Height S: Energy Analysis
The problem asks for the height from the surface and the speed of a particle at a specific instant, given its initial release height and a relationship between its kinetic and potential energies.
Applying Conservation of Energy
Let the initial height be $S$. The initial potential energy (PE) is $PE_i = mgS$, and the initial kinetic energy (KE) is $KE_i = 0$ (since released from rest).
At a certain height $h$ from the surface, the potential energy is $PE_f = mgh$. The kinetic energy is $KE_f$. The total energy at this point is $E_f = PE_f + KE_f = mgh + KE_f$.
By the conservation of mechanical energy, the total energy remains constant:
$E_i = E_f$
$mgS + 0 = mgh + KE_f$
$mgS = mgh + KE_f$
Using the Energy Relationship
We are given that at height $h$, the kinetic energy is three times the potential energy:
$KE_f = 3 \times PE_f$
$KE_f = 3mgh$
Calculating the Height
Substitute the expression for $KE_f$ back into the conservation of energy equation:
$mgS = mgh + 3mgh$
$mgS = 4mgh$
Dividing both sides by $mg$ (assuming $m \neq 0$ and $g \neq 0$), we get:
$S = 4h$
Therefore, the height $h$ from the surface is:
$h = \frac{S}{4}$
Calculating the Speed
Now we need to find the speed $v$ at this height $h$. We know that $KE_f = \frac{1}{2}mv^2$. Using the relationship $KE_f = 3mgh$:
$\frac{1}{2}mv^2 = 3mgh$
Substitute the calculated height $h = \frac{S}{4}$:
$\frac{1}{2}mv^2 = 3mg\left(\frac{S}{4}\right)$
$\frac{1}{2}mv^2 = \frac{3mgS}{4}$
Cancel mass $m$ from both sides:
$\frac{1}{2}v^2 = \frac{3gS}{4}$
Solve for $v^2$:
$v^2 = 2 \times \frac{3gS}{4}$
$v^2 = \frac{3gS}{2}$
Taking the square root to find the speed $v$:
$v = \sqrt{\frac{3gS}{2}}$
Final Answer
The height from the surface is $\frac{S}{4}$ and the speed is $\sqrt{\frac{3gS}{2}}$.
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