To solve this problem, we need to determine from which angle \(\theta\) the ram should be released so that it strikes the object P with a horizontal velocity of 5 m/s.

First, we apply the principle of conservation of mechanical energy. The potential energy at the height when the ram is released will convert to kinetic energy at the lowest point (where the ram strikes P).

1. Consider the initial position (at an angle \(\theta\)):
   • Height of the center of mass from the lowest point: \(h = L - L \cos \theta\), where \(L = 7 \text{ m}\) (length of rope).
   • Initial potential energy: \(PE = mgh = mg(L - L\cos\theta)\).
2. At the lowest position (just before striking P):
   • All potential energy is converted to kinetic energy.
   • Kinetic energy: \(KE = \frac{1}{2}mv^2\), where \(v = 5 \text{ m/s}\).

Since initial potential energy equals kinetic energy at the lowest point, we have:

mg(L - L \cos \theta) = \frac{1}{2}mv^2

Cancel out the common term \(m\) from both sides, we get:

g(L - L \cos \theta) = \frac{1}{2}v^2

Plug in the given values:

• \(g = 9.81 \text{ m/s}^2\)
• \(v = 5 \text{ m/s}\)
• \(L = 7 \text{ m}\)

Thus, we have:

9.81 \times (7 - 7\cos\theta) = \frac{1}{2} \times 5^2

Simplify the right-hand side:

9.81 \times 7(1 - \cos\theta) = 12.5

Rearrange to solve for \(\cos\theta\):

\cos\theta = 1 - \frac{12.5}{9.81 \times 7}

Calculate:

\cos\theta = 1 - \frac{12.5}{68.67} = 1 - 0.182

\(\cos\theta \approx 0.818\)

Finally, calculate \(\theta\):

\theta = \cos^{-1}(0.818) \approx 35.1^\circ

Therefore, the correct option is 35.1°.