A ball is thrown with velocity `8 ms^(-1)` making an angle `60^(@)` with the horizontal.Its velocity will be perpendicular to the direction of initial velocity of projection after a time of

Options

A. `1.6/sqrt3s`
B. `4/sqrt3s`
C. `0.6 s`
D. `1.6sqrt3 s`

Explanation

To solve the problem, we need to determine the time at which the velocity of the ball becomes perpendicular to its initial velocity. Here’s a step-by-step solution:

### Step 1: Understand the initial conditions
The ball is thrown with an initial velocity \( u = 8 \, \text{m/s} \) at an angle \( \theta = 60^\circ \) with the horizontal.

### Step 2: Break down the initial velocity into components
The initial velocity can be broken down into horizontal and vertical components:
- Horizontal component:
\[ u_x = u \cos \theta = 8 \cos(60^\circ) = 8 \times \frac{1}{2} = 4 \, \text{m/s} \]
- Vertical component:
\[ u_y = u \sin \theta = 8 \sin(60^\circ) = 8 \times \frac{\sqrt{3}}{2} = 4\sqrt{3} \, \text{m/s} \]

### Step 3: Write the equations for velocity after time \( t \)
The velocity of the ball at time \( t \) can be expressed as:
- Horizontal velocity remains constant:
\[ v_x = u_x = 4 \, \text{m/s} \]
- Vertical velocity changes due to gravity:
\[ v_y = u_y - gt = 4\sqrt{3} - 10t \, \text{m/s} \]

### Step 4: Condition for perpendicularity
The velocities will be perpendicular when the dot product of the initial velocity vector and the velocity vector at time \( t \) is zero. The initial velocity vector is:
\[ \vec{u} = (u_x, u_y) = (4, 4\sqrt{3}) \]
The velocity vector at time \( t \) is:
\[ \vec{v} = (v_x, v_y) = (4, 4\sqrt{3} - 10t) \]
Setting the dot product to zero:
\[ \vec{u} \cdot \vec{v} = 4 \cdot 4 + 4\sqrt{3} \cdot (4\sqrt{3} - 10t) = 0 \]
Expanding this:
\[ 16 + 12(4\sqrt{3}) - 40\sqrt{3}t = 0 \]
This simplifies to:
\[ 16 + 12\sqrt{3} - 40\sqrt{3}t = 0 \]

### Step 5: Solve for \( t \)
Rearranging the equation gives:
\[ 40\sqrt{3}t = 16 + 12\sqrt{3} \]
\[ t = \frac{16 + 12\sqrt{3}}{40\sqrt{3}} \]

### Step 6: Simplify the expression for \( t \)
Calculating the numerical value:
\[ t = \frac{16}{40\sqrt{3}} + \frac{12\sqrt{3}}{40\sqrt{3}} = \frac{16 + 12}{40\sqrt{3}} = \frac{28}{40\sqrt{3}} = \frac{7}{10\sqrt{3}} \approx 0.404 \, \text{s} \]

### Final Answer
The time after which the velocity of the ball will be perpendicular to the direction of the initial velocity of projection is approximately \( t \approx 0.404 \, \text{s} \).

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