Two particles were projected one by one with the same initial velocity from the same point on level ground. They follow the same parabolic trajectory and are found to be in the same horizontal level, separated by a distance of `1 m, 2 s` after the second partice was projected. Assume that the horizontal component of their velocities is `0.5 ms^-1`. Find
(a) the horizontal range of the parabolic path.
(b) the maximum height for the parabolic path.

Explanation

To solve the problem, we need to find the horizontal range and the maximum height of the parabolic trajectory of the two particles projected from the same point.

### Step-by-Step Solution:

**Step 1: Understand the problem setup.**
- Two particles are projected one after the other with the same initial velocity.
- The horizontal component of their velocities is given as \(0.5 \, \text{m/s}\).
- They are separated by a distance of \(1 \, \text{m}\) after \(2 \, \text{s}\) from the time the second particle is projected.

**Step 2: Calculate the horizontal distance traveled by the second particle.**
- The second particle travels for \(2 \, \text{s}\) before the first particle reaches the same horizontal level.
- The distance traveled by the second particle in \(2 \, \text{s}\) is:
\[ \text{Distance} = \text{Velocity} \times \text{Time} = 0.5 \, \text{m/s} \times 2 \, \text{s} = 1 \, \text{m} \]

**Step 3: Determine the total horizontal distance (horizontal range) covered by the first particle.**
- The first particle travels for a total of \(4 \, \text{s}\) (2 seconds before the second particle is projected and 2 seconds after).
- The distance traveled by the first particle in \(4 \, \text{s}\) is:
\[ \text{Distance} = 0.5 \, \text{m/s} \times 4 \, \text{s} = 2 \, \text{m} \]
- Adding the \(1 \, \text{m}\) distance (the distance between the two particles when the second is projected), the total horizontal range is:
\[ \text{Total Distance} = 1 \, \text{m} + 2 \, \text{m} = 3 \, \text{m} \]

**Step 4: Calculate the time of flight for the particles.**
- The total time of flight for the first particle is \(6 \, \text{s}\) (2 seconds before the second particle is projected + 4 seconds of flight after).

**Step 5: Use the time of flight to find the vertical component of the initial velocity.**
- The time of flight \(T\) is given by the formula:
\[ T = \frac{2u \sin \theta}{g} \]
where \(g = 10 \, \text{m/s}^2\).
- Rearranging gives:
\[ u \sin \theta = \frac{gT}{2} = \frac{10 \times 6}{2} = 30 \, \text{m/s} \]

**Step 6: Calculate the maximum height reached by the particles.**
- The maximum height \(H\) is given by:
\[ H = \frac{u^2 \sin^2 \theta}{2g} \]
- We need to find \(u\) first. From the horizontal motion, we know:
\[ u \cos \theta = 0.5 \, \text{m/s} \]
- Using the Pythagorean theorem for the components of velocity:
\[ u^2 = (u \sin \theta)^2 + (u \cos \theta)^2 \]
Substituting \(u \sin \theta = 30\) and \(u \cos \theta = 0.5\):
\[ u^2 = 30^2 + 0.5^2 = 900 + 0.25 = 900.25 \]
Thus,
\[ u = \sqrt{900.25} \approx 30 \, \text{m/s} \]

- Now substituting \(u\) back to find \(H\):
\[ H = \frac{(30)^2 \sin^2 \theta}{2 \times 10} \]
Since we have \(u \sin \theta = 30\), we can find \(\sin^2 \theta\):
\[ H = \frac{900 \sin^2 \theta}{20} = 45 \, \text{m} \]

### Final Answers:
(a) The horizontal range of the parabolic path is \(3 \, \text{m}\).
(b) The maximum height for the parabolic path is \(45 \, \text{m}\).