A particle is projected with a velocity of `39.2 m//"sec"` at an angle of `30^(@)` to the horizontal. It will move at right angles to the direction of projection after the time

Options

A. 8 sec
B. 5 sec
C. 6 sec
D. 10 sec

Answer

To solve the problem of a particle projected with a velocity of `39.2 m/s` at an angle of `30°` to the horizontal, we need to determine the time at which the particle's velocity becomes perpendicular to its initial velocity. Here’s a step-by-step solution:

### Step 1: Break down the initial velocity into components
The initial velocity \( u \) can be broken down into horizontal and vertical components using trigonometric functions:
- \( u_x = u \cos(\theta) \)
- \( u_y = u \sin(\theta) \)

Given:
- \( u = 39.2 \, \text{m/s} \)
- \( \theta = 30° \)

Calculating the components:
- \( u_x = 39.2 \cos(30°) = 39.2 \times \frac{\sqrt{3}}{2} = 19.6 \sqrt{3} \, \text{m/s} \)
- \( u_y = 39.2 \sin(30°) = 39.2 \times \frac{1}{2} = 19.6 \, \text{m/s} \)

### Step 2: Determine the velocity components at time \( t \)
The horizontal component of velocity remains constant:
- \( v_x = u_x = 19.6 \sqrt{3} \, \text{m/s} \)

The vertical component of velocity changes due to gravity:
- \( v_y = u_y - g t \)
Where \( g = 9.8 \, \text{m/s}^2 \).

Thus,
- \( v_y = 19.6 - 9.8 t \)

### Step 3: Set up the condition for perpendicularity
For the two velocity vectors to be perpendicular, their dot product must equal zero:
\[ u_x v_x + u_y v_y = 0 \]

Substituting the known values:
\[ (19.6 \sqrt{3})(19.6 \sqrt{3}) + (19.6)(19.6 - 9.8 t) = 0 \]

### Step 4: Simplify the equation
Expanding the equation gives:
\[ 19.6^2 \cdot 3 + 19.6^2 - 19.6 \cdot 9.8 t = 0 \]

Combining like terms:
\[ 19.6^2 (3 + 1) - 19.6 \cdot 9.8 t = 0 \]
\[ 19.6^2 \cdot 4 - 19.6 \cdot 9.8 t = 0 \]

### Step 5: Solve for \( t \)
Factoring out \( 19.6 \):
\[ 19.6 (19.6 \cdot 4 - 9.8 t) = 0 \]

Since \( 19.6 \neq 0 \), we can set the remaining factor to zero:
\[ 19.6 \cdot 4 - 9.8 t = 0 \]
\[ 9.8 t = 19.6 \cdot 4 \]
\[ t = \frac{19.6 \cdot 4}{9.8} \]
\[ t = \frac{78.4}{9.8} = 8 \, \text{s} \]

### Final Answer
The time at which the particle moves at right angles to the direction of projection is **8 seconds**.