A projectile is thrown with a velocity of 20 `m//s,` at an angle of `60^(@)` with the horizontal. After how much time the velocity vector will make an angle of `45^(@)` with the horizontal (in upward direction) is (take g=`10m//s^(2))-`

Options

A. `sqrt3s`
B. `(1)/(sqrt3)s`
C. `(sqrt3+1)s`
D. `(sqrt3-1)s`

Answer

To solve the problem, we need to determine the time at which the velocity vector of a projectile makes an angle of 45 degrees with the horizontal. The projectile is launched with an initial velocity of 20 m/s at an angle of 60 degrees. We will use the following steps:

### Step 1: Determine the initial velocity components
The initial velocity \( V_0 \) can be broken down into its horizontal and vertical components using trigonometric functions.

- Horizontal component \( V_{0x} = V_0 \cos(60^\circ) \)
- Vertical component \( V_{0y} = V_0 \sin(60^\circ) \)

Given \( V_0 = 20 \, \text{m/s} \):
- \( V_{0x} = 20 \cos(60^\circ) = 20 \times \frac{1}{2} = 10 \, \text{m/s} \)
- \( V_{0y} = 20 \sin(60^\circ) = 20 \times \frac{\sqrt{3}}{2} = 10\sqrt{3} \, \text{m/s} \)

### Step 2: Write the equations for velocity components at time \( t \)
The horizontal component of velocity remains constant since there is no horizontal acceleration:
- \( V_x = V_{0x} = 10 \, \text{m/s} \)

The vertical component of velocity changes due to gravity:
- \( V_y = V_{0y} - g t \)
- \( V_y = 10\sqrt{3} - 10t \)

### Step 3: Set the condition for the angle of 45 degrees
For the velocity vector to make an angle of 45 degrees with the horizontal, the magnitudes of the horizontal and vertical components of the velocity must be equal:
- \( V_x = V_y \)

Substituting the values:
- \( 10 = 10\sqrt{3} - 10t \)

### Step 4: Solve for time \( t \)
Rearranging the equation:
- \( 10t = 10\sqrt{3} - 10 \)
- \( t = \sqrt{3} - 1 \)

### Step 5: Calculate the time
Using the approximate value of \( \sqrt{3} \approx 1.732 \):
- \( t \approx 1.732 - 1 = 0.732 \, \text{s} \)

### Final Answer
The time after which the velocity vector will make an angle of 45 degrees with the horizontal is approximately \( 0.732 \, \text{s} \).

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