The position coordinates of a projectile projected from ground on a certain planet (with an atmosphere) are given by `y = (4t – 2t^(2))` m and `x = (3t)` metre, where t is in second and point of projection is taken as origin. The angle of projection of projectile with vertical is -

Options

A. `30^(@)`
B. `37^(@)`
C. `45^(@)`
D. `60^(@)`

Answer

To find the angle of projection of a projectile with respect to the vertical, we can follow these steps:

### Step-by-Step Solution:

1. **Identify the equations of motion**: The position coordinates of the projectile are given as:
\[ y = 4t - 2t^2 \quad \text{(1)} \]
\[ x = 3t \quad \text{(2)} \]

2. **Find the velocity components**: The velocity components can be obtained by differentiating the position equations with respect to time \( t \).

- For the vertical component \( v_y \):
\[ v_y = \frac{dy}{dt} = \frac{d}{dt}(4t - 2t^2) = 4 - 4t \quad \text{(3)} \]

- For the horizontal component \( v_x \):
\[ v_x = \frac{dx}{dt} = \frac{d}{dt}(3t) = 3 \quad \text{(4)} \]

3. **Evaluate the velocities at \( t = 0 \)**: At the moment of projection (\( t = 0 \)):
- From equation (3):
\[ v_y = 4 - 4(0) = 4 \, \text{m/s} \]
- From equation (4):
\[ v_x = 3 \, \text{m/s} \]

4. **Calculate the tangent of the angle of projection**: The angle \( \theta \) with respect to the vertical can be found using the tangent function:
\[ \tan(\theta) = \frac{v_x}{v_y} = \frac{3}{4} \quad \text{(5)} \]

5. **Determine the angle \( \theta \)**: To find \( \theta \), we can use the arctangent function:
\[ \theta = \tan^{-1}\left(\frac{3}{4}\right) \]

6. **Calculate the angle in degrees**: Using a calculator or trigonometric tables:
\[ \theta \approx 36.87^\circ \approx 37^\circ \]

### Final Answer:
The angle of projection of the projectile with respect to the vertical is approximately \( 37^\circ \).

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