A projectile of mass `1kg` is projected with a velocity of `sqrt(20)m//s` such that it strikes on the same level as the point of projection at a distance of `sqrt(3)m`. Which of the following options are incorrect:

Options

A. The maximum height reached by the projectile can be 0.25 m.
B. The minimum velocity during its motion can be `sqrt()5` m/s.
C. The time taken for the flight can be `sqrt(3//5) s`.
D. Minimum kinetic energy during its motion can be 6 J.

Answer

To solve the problem step by step, we will analyze the projectile motion of the object and determine which of the provided options are incorrect.

### Step 1: Identify Given Data
- Mass of the projectile, \( m = 1 \, \text{kg} \)
- Initial velocity, \( u = \sqrt{20} \, \text{m/s} \)
- Range, \( R = \sqrt{3} \, \text{m} \)
- Acceleration due to gravity, \( g = 10 \, \text{m/s}^2 \)

### Step 2: Use the Range Formula
The range \( R \) of a projectile is given by the formula:
\[ R = \frac{u^2 \sin(2\theta)}{g} \]
Substituting the known values:
\[ \sqrt{3} = \frac{(\sqrt{20})^2 \sin(2\theta)}{10} \]
This simplifies to:
\[ \sqrt{3} = \frac{20 \sin(2\theta)}{10} \]
\[ \sqrt{3} = 2 \sin(2\theta) \]
\[ \sin(2\theta) = \frac{\sqrt{3}}{2} \]

### Step 3: Determine Possible Angles
The values of \( 2\theta \) that satisfy \( \sin(2\theta) = \frac{\sqrt{3}}{2} \) are:
\[ 2\theta = 60^\circ \quad \text{or} \quad 2\theta = 120^\circ \]
Thus, the possible angles \( \theta \) are:
\[ \theta = 30^\circ \quad \text{or} \quad \theta = 60^\circ \]

### Step 4: Calculate Maximum Height
The maximum height \( H \) of the projectile is given by:
\[ H = \frac{u^2 \sin^2(\theta)}{2g} \]

**For \( \theta = 30^\circ \)**:
\[ H = \frac{(\sqrt{20})^2 \sin^2(30^\circ)}{2 \times 10} = \frac{20 \times \left(\frac{1}{2}\right)^2}{20} = \frac{20 \times \frac{1}{4}}{20} = 0.25 \, \text{m} \]

**For \( \theta = 60^\circ \)**:
\[ H = \frac{(\sqrt{20})^2 \sin^2(60^\circ)}{2 \times 10} = \frac{20 \times \left(\frac{\sqrt{3}}{2}\right)^2}{20} = \frac{20 \times \frac{3}{4}}{20} = 0.75 \, \text{m} \]

### Step 5: Analyze Options
Now we need to analyze the options given in the question:

1. **Maximum height can be 0.5 m**: This is incorrect since the maximum height is 0.75 m for \( \theta = 60^\circ \) and 0.25 m for \( \theta = 30^\circ \).

2. **Minimum velocity during its motion can be \( \sqrt{5} \, \text{m/s} \)**: This is incorrect. The minimum velocity occurs at the maximum height where the vertical component is zero, and the horizontal component is \( u \cos(\theta) \). For \( \theta = 30^\circ \), \( u \cos(30^\circ) = \sqrt{20} \cdot \frac{\sqrt{3}}{2} = \sqrt{15} \, \text{m/s} \), which is greater than \( \sqrt{5} \, \text{m/s} \).

3. **Time taken for the flight can be \( \frac{\sqrt{3}}{5} \, \text{s}**: This is correct. The time of flight can be calculated using \( t = \frac{2u \sin(\theta)}{g} \). For both angles, this value can be verified.

4. **Minimum kinetic energy during its motion can be 6 Joules**: This is incorrect. The minimum kinetic energy occurs at the maximum height, and can be calculated as \( KE = \frac{1}{2} m v_x^2 \) where \( v_x = u \cos(\theta) \). For \( \theta = 30^\circ \), this gives a minimum kinetic energy of approximately 2 Joules.

### Conclusion
The incorrect options are:
1. Maximum height can be 0.5 m.
2. Minimum velocity during its motion can be \( \sqrt{5} \, \text{m/s} \).
3. Minimum kinetic energy during its motion can be 6 Joules.