A parojectile of mass `1kg` is projected with a velcoity 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 `0.25m`.
B. the minimum velocity during its motion can be `sqrt(15)m//s`
C. the time taken for the flight can be `sqrt((3)/(5))sec`.
D. maximum potential enegry during its motion can be `6J`.

Explanation

To solve the problem, we need to analyze the projectile motion of the given object and determine which of the provided options are incorrect. Here’s a step-by-step breakdown of the solution:

### Step 1: Understand the Given Information
- Mass of the projectile (m) = 1 kg
- Initial velocity (u) = √20 m/s
- Range (R) = √3 m
- Acceleration due to gravity (g) = 10 m/s²

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

### Step 3: Find Possible Angles
From the sine value, we can find:
\[ 2\theta = 60^\circ \text{ or } 120^\circ \]
Thus,
\[ \theta = 30^\circ \text{ or } 60^\circ \]

### Step 4: Calculate Maximum Height
Using the formula for maximum height \( H \):
\[ H = \frac{u^2 \sin^2(\theta)}{2g} \]

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

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

### Step 5: Calculate Time of Flight
The time of flight \( T \) is given by:
\[ T = \frac{2u \sin(\theta)}{g} \]

**For \( \theta = 30^\circ \)**:
\[ T = \frac{2 \times \sqrt{20} \times \sin(30^\circ)}{10} \]
\[ T = \frac{2 \times \sqrt{20} \times \frac{1}{2}}{10} = \frac{\sqrt{20}}{10} = \frac{\sqrt{20}}{10} = \frac{2\sqrt{5}}{10} = \frac{\sqrt{5}}{5} \text{ seconds} \]

**For \( \theta = 60^\circ \)**:
\[ T = \frac{2 \times \sqrt{20} \times \sin(60^\circ)}{10} \]
\[ T = \frac{2 \times \sqrt{20} \times \frac{\sqrt{3}}{2}}{10} = \frac{\sqrt{20} \sqrt{3}}{10} = \frac{\sqrt{60}}{10} = \frac{\sqrt{15}}{5} \text{ seconds} \]

### Step 6: Calculate Maximum Potential Energy
The maximum potential energy (PE) at the maximum height is given by:
\[ PE = mgh \]
For \( \theta = 30^\circ \):
\[ PE = 1 \times 10 \times 0.25 = 2.5 \text{ joules} \]

### Step 7: Evaluate the Options
1. Maximum height raised by the projectile can be 0.25 meters. **(Correct)**
2. Minimum velocity during its motion can be \( \sqrt{15} \) m/s. **(Incorrect)**
3. Time taken for the flight can be \( \frac{\sqrt{3}}{5} \) seconds. **(Correct)**
4. Maximum potential energy during its motion can be 6 joules. **(Incorrect)**

### Conclusion
The incorrect options are:
- Minimum velocity during its motion can be \( \sqrt{15} \) m/s.
- Maximum potential energy during its motion can be 6 joules.