To find the potential energy at the highest point of the trajectory of a body projected at an angle θ to the horizontal with kinetic energy Ek, we can follow these steps:
### Step-by-Step Solution:
1. **Understanding the Components of Velocity:**
- When a body is projected at an angle θ, its initial velocity (u) can be broken down into horizontal and vertical components:
- Horizontal component: \( u_x = u \cos \theta \)
- Vertical component: \( u_y = u \sin \theta \)
2. **Kinetic Energy at the Initial Point:**
- The initial kinetic energy (Ek) is given by:
\[
E_k = \frac{1}{2} m u^2
\]
3. **Conservation of Mechanical Energy:**
- As the body moves upward, the mechanical energy is conserved (assuming no air resistance). Thus, the sum of kinetic energy and potential energy at the initial point equals the sum at the highest point:
\[
E_k + 0 = E_{k, \text{final}} + E_{p, \text{final}}
\]
4. **Kinetic Energy at the Highest Point:**
- At the highest point of the trajectory, the vertical component of velocity becomes zero (Vy = 0), and only the horizontal component remains:
- \( u_x = u \cos \theta \)
- Therefore, the kinetic energy at the highest point is:
\[
E_{k, \text{final}} = \frac{1}{2} m (u \cos \theta)^2 = \frac{1}{2} m u^2 \cos^2 \theta
\]
5. **Relating Final Kinetic Energy to Initial Kinetic Energy:**
- Since \( E_k = \frac{1}{2} m u^2 \), we can express the final kinetic energy in terms of Ek:
\[
E_{k, \text{final}} = E_k \cos^2 \theta
\]
6. **Calculating Potential Energy at the Highest Point:**
- Using conservation of energy:
\[
E_k = E_{k, \text{final}} + E_{p, \text{final}}
\]
- Rearranging gives us:
\[
E_{p, \text{final}} = E_k - E_{k, \text{final}} = E_k - E_k \cos^2 \theta
\]
- Factoring out \( E_k \):
\[
E_{p, \text{final}} = E_k (1 - \cos^2 \theta)
\]
- Using the identity \( 1 - \cos^2 \theta = \sin^2 \theta \):
\[
E_{p, \text{final}} = E_k \sin^2 \theta
\]
7. **Final Answer:**
- The potential energy at the highest point of the trajectory is:
\[
E_{p, \text{final}} = E_k \sin^2 \theta
\]
