A projectile of mass `2kg` has velocities `3m//s` and `4m/s` at two points during its flight in the uniform gravitational field of the earth. If these two velocities are perpendicular to each other, then the minimum kinetic enerty of the particle during its flight is

Options

A. 6.32 J
B. 8.40 J
C. 16.32 J
D. 5.76 J

Explanation

To solve the problem, we need to find the minimum kinetic energy of a projectile given that it has two velocities that are perpendicular to each other. Let's break down the solution step by step.

### Step 1: Understand the Given Information
We have a projectile of mass \( m = 2 \, \text{kg} \) with two velocities:
- \( v_1 = 3 \, \text{m/s} \)
- \( v_2 = 4 \, \text{m/s} \)

These velocities are perpendicular to each other.

### Step 2: Use the Relationship Between Velocities
Since the velocities are perpendicular, we can use the relationship of their components. Let's denote the angle of \( v_1 \) with the horizontal as \( \alpha \). The angle of \( v_2 \) will then be \( 90^\circ - \alpha \).

The horizontal components of the velocities must be equal because there is no horizontal acceleration in projectile motion:
\[ v_1 \cos(\alpha) = v_2 \cos(90^\circ - \alpha) = v_2 \sin(\alpha) \]
Substituting the values:
\[ 3 \cos(\alpha) = 4 \sin(\alpha) \]

### Step 3: Solve for \( \tan(\alpha) \)
Rearranging the equation gives:
\[ \frac{\cos(\alpha)}{\sin(\alpha)} = \frac{4}{3} \]
Thus,
\[ \tan(\alpha) = \frac{3}{4} \]

### Step 4: Find \( \cos(\alpha) \) and \( \sin(\alpha) \)
Using the identity \( \tan(\alpha) = \frac{\text{opposite}}{\text{adjacent}} \):
- Let the opposite side be 3 and the adjacent side be 4.
- The hypotenuse \( h \) can be calculated using Pythagoras' theorem:
\[ h = \sqrt{3^2 + 4^2} = \sqrt{9 + 16} = 5 \]
Thus,
\[ \sin(\alpha) = \frac{3}{5}, \quad \cos(\alpha) = \frac{4}{5} \]

### Step 5: Calculate \( u \cos(\theta) \)
The minimum velocity occurs at the highest point of the projectile's motion, which is given by:
\[ u \cos(\theta) = v_1 \cos(\alpha) = 3 \cdot \frac{4}{5} = \frac{12}{5} \, \text{m/s} \]

### Step 6: Calculate Minimum Kinetic Energy
The kinetic energy \( KE \) is given by:
\[ KE = \frac{1}{2} m (u \cos(\theta))^2 \]
Substituting the values:
\[ KE = \frac{1}{2} \cdot 2 \cdot \left(\frac{12}{5}\right)^2 \]
\[ KE = 1 \cdot \frac{144}{25} = \frac{144}{25} \, \text{J} \]

### Step 7: Convert to Decimal Form
To express this in decimal form:
\[ \frac{144}{25} = 5.76 \, \text{J} \]

### Final Answer
Thus, the minimum kinetic energy of the projectile during its flight is:
\[ \boxed{5.76 \, \text{J}} \]