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 the projectile with the vertical, we can follow these steps:

### Step 1: Identify the equations of motion
The position coordinates of the projectile are given as:
- \( y = 4t - 2t^2 \) (vertical motion)
- \( x = 3t \) (horizontal motion)

### Step 2: Differentiate the equations to find velocity components
To find the velocity components, we differentiate the position equations with respect to time \( t \).

- For the horizontal motion:
\[ v_x = \frac{dx}{dt} = \frac{d(3t)}{dt} = 3 \, \text{m/s} \]

- For the vertical motion:
\[ v_y = \frac{dy}{dt} = \frac{d(4t - 2t^2)}{dt} = 4 - 4t \, \text{m/s} \]

### Step 3: Calculate the initial velocity components
At \( t = 0 \):
- The initial horizontal component of velocity \( v_x = 3 \, \text{m/s} \)
- The initial vertical component of velocity \( v_y = 4 - 4(0) = 4 \, \text{m/s} \)

### Step 4: Determine the angle of projection
The angle \( \theta \) with respect to the vertical can be found using the tangent function:
\[ \tan(\theta) = \frac{v_x}{v_y} \]
Substituting the values:
\[ \tan(\theta) = \frac{3}{4} \]

### Step 5: Calculate the angle using the arctangent function
To find \( \theta \):
\[ \theta = \tan^{-1}\left(\frac{3}{4}\right) \]

Using a calculator or trigonometric tables:
\[ \theta \approx 36.87^\circ \approx 37^\circ \]

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

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