A particle is projected upwards with a velocity of `100 m//s` at an angle of `60^(@)` with the vertical. Find the time when the particle will move perpendicular to its initial direction, taking `g = 10 m//s^(2)`.

Answer

To solve the problem step by step, we will analyze the motion of the particle projected at an angle with respect to the vertical.

### Step 1: Understand the Components of Velocity
The initial velocity \( u \) of the particle is given as \( 100 \, \text{m/s} \) at an angle of \( 60^\circ \) with the vertical. We need to find the horizontal and vertical components of this velocity.

- The horizontal component \( u_x \) (along the x-axis) is given by:
\[ u_x = u \sin(60^\circ) = 100 \sin(60^\circ) = 100 \cdot \frac{\sqrt{3}}{2} = 50\sqrt{3} \, \text{m/s} \]

- The vertical component \( u_y \) (along the y-axis) is given by:
\[ u_y = u \cos(60^\circ) = 100 \cos(60^\circ) = 100 \cdot \frac{1}{2} = 50 \, \text{m/s} \]

### Step 2: Determine the Condition for Perpendicular Motion
The problem states that we need to find the time when the particle's motion becomes perpendicular to its initial direction. For two vectors to be perpendicular, their dot product must equal zero.

Let \( v_x \) and \( v_y \) be the horizontal and vertical components of the velocity at time \( t \).

### Step 3: Express the Components of Velocity
Since there is no acceleration in the x-direction, the horizontal velocity remains constant:
\[ v_x = u_x = 50\sqrt{3} \, \text{m/s} \]

For the vertical component, the velocity changes due to gravity:
\[ v_y = u_y - gt = 50 - 10t \]

### Step 4: Set Up the Dot Product Condition
The dot product of the initial velocity vector \( \mathbf{u} = (u_x, u_y) \) and the final velocity vector \( \mathbf{v} = (v_x, v_y) \) must be zero:
\[ u_x v_x + u_y v_y = 0 \]

Substituting the values:
\[ (50\sqrt{3})(50\sqrt{3}) + (50)(50 - 10t) = 0 \]

### Step 5: Simplify the Equation
Calculating the first term:
\[ (50\sqrt{3})^2 = 2500 \cdot 3 = 7500 \]
So, the equation becomes:
\[ 7500 + 2500 - 500t = 0 \]
\[ 10000 - 500t = 0 \]

### Step 6: Solve for Time \( t \)
Rearranging the equation gives:
\[ 500t = 10000 \]
\[ t = \frac{10000}{500} = 20 \, \text{s} \]

### Final Answer
The time when the particle will move perpendicular to its initial direction is \( t = 20 \, \text{s} \).

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