Let's solve this problem step by step. We'll find out how far beyond the wall the stone will strike the ground.
Given data:
- Initial velocity, \(u = 50 \, \text{m/s}\)
- Angle of projection, \(\theta = 30^\circ\)
- Time to cross the wall, \(t = 3 \, \text{s}\)
- Acceleration due to gravity, \(g = 10 \, \text{m/s}^2\)
We will use the horizontal and vertical components of the initial velocity:
- Horizontal component, \(u_x = u \cos \theta = 50 \cos 30^\circ\)
- Vertical component, \(u_y = u \sin \theta = 50 \sin 30^\circ\)
Calculate these components:
- \(u_x = 50 \times \frac{\sqrt{3}}{2} = 25\sqrt{3} \, \text{m/s}\)
- \(u_y = 50 \times \frac{1}{2} = 25 \, \text{m/s}\)
Now, calculate the horizontal distance covered in 3 seconds:
- \(x = u_x \cdot t = 25\sqrt{3} \times 3 = 75\sqrt{3} \, \text{m}\)
Calculate the total time of flight. We use the formula for vertical motion (ignoring air resistance):
- Time of flight, \(T = \frac{2u_y}{g} = \frac{2 \times 25}{10} = 5 \, \text{s}\)
Now, calculate the total horizontal distance (range) covered before hitting the ground:
- Total horizontal range, \(R = u_x \cdot T = 25\sqrt{3} \times 5 = 125\sqrt{3} \, \text{m}\)
To find how far beyond the wall the stone will strike the ground, subtract the distance to the wall from the total range:
- Distance beyond the wall, \(= R - x = 125\sqrt{3} - 75\sqrt{3} = 50\sqrt{3} \, \text{m}\)
Simplifying it further:
- Calculate \(50\sqrt{3} \approx 50 \times 1.732 = 86.6 \, \text{m}\)
Hence, the stone will strike the ground approximately 86.5 m beyond the wall. Thus, the correct option is Option C: 86.5 m.