To solve the problem step by step, we start with the information given in the question:
1. **Understanding the Problem**:
- A body is projected at an angle of \(60^\circ\) with the horizontal.
- The kinetic energy (KE) at its maximum height is given as \(10 \, \text{J}\).
- We need to find the height at which the potential energy (PE) equals the kinetic energy.
2. **Finding the Initial Kinetic Energy**:
- At the maximum height, the vertical component of the velocity becomes zero, and only the horizontal component remains.
- The horizontal component of the initial velocity can be expressed as:
\[
v_{\text{top}} = u \cos(60^\circ)
\]
- The kinetic energy at the maximum height is given by:
\[
KE = \frac{1}{2} m v_{\text{top}}^2 = 10 \, \text{J}
\]
- Substituting \(v_{\text{top}}\):
\[
\frac{1}{2} m (u \cos(60^\circ))^2 = 10
\]
- Since \(\cos(60^\circ) = \frac{1}{2}\):
\[
\frac{1}{2} m \left(u \cdot \frac{1}{2}\right)^2 = 10
\]
- This simplifies to:
\[
\frac{1}{2} m \cdot \frac{u^2}{4} = 10
\]
- Rearranging gives:
\[
\frac{m u^2}{8} = 10 \implies m u^2 = 80 \implies u^2 = \frac{80}{m}
\]
3. **Total Energy of the System**:
- The total energy (E) at the point of projection is entirely kinetic:
\[
E = KE_{\text{initial}} = \frac{1}{2} m u^2 = 40 \, \text{J}
\]
4. **Finding Maximum Potential Energy**:
- At maximum height, the potential energy (PE) is maximum and is given by:
\[
PE_{\text{max}} = E - KE_{\text{max}} = 40 \, \text{J} - 10 \, \text{J} = 30 \, \text{J}
\]
5. **Finding the Height Where PE = KE**:
- We need to find the height \(h\) where PE equals KE. Let \(PE = KE = x\).
- Therefore, we have:
\[
PE + KE = 40 \implies x + x = 40 \implies 2x = 40 \implies x = 20 \, \text{J}
\]
- Thus, at the height where PE equals KE, both are \(20 \, \text{J}\).
6. **Relating Potential Energy to Height**:
- The potential energy at height \(h\) is given by:
\[
PE = mgh
\]
- Setting this equal to \(20 \, \text{J}\):
\[
mgh = 20
\]
7. **Finding the Maximum Height**:
- The maximum potential energy (at maximum height) is:
\[
PE_{\text{max}} = mgh_{\text{max}} = 30 \, \text{J}
\]
- Thus, we can express the maximum height \(h_{\text{max}}\) as:
\[
mgh_{\text{max}} = 30 \implies h_{\text{max}} = \frac{30}{mg}
\]
8. **Finding the Ratio of Heights**:
- The ratio of the height where PE = KE to the maximum height is:
\[
\frac{h}{h_{\text{max}}} = \frac{\frac{20}{mg}}{\frac{30}{mg}} = \frac{20}{30} = \frac{2}{3}
\]
9. **Final Height Calculation**:
- Therefore, the height at which the potential energy and kinetic energy are equal is:
\[
h = \frac{2}{3} h_{\text{max}}
\]
### Conclusion:
The height at which the potential energy and kinetic energy are equal is \(\frac{2}{3}\) of the maximum height.
