In a projectile motion, the height `y = sqrt(3)t - 5t^(2) + r^(3)` and horizontal distance `x = t + 2t^(2) -t^(3)` . The angle of projection is given by

Options

A. `30^(@)`
B. `60^(@)`
C. `45^(@)`
D. `75^(@)`

Explanation

To find the angle of projection in the given projectile motion, we will follow these steps:

### Step 1: Differentiate the height equation
The height \( y \) is given by:
\[ y = \sqrt{3}t - 5t^2 + t^3 \]
We need to find the vertical velocity \( V_y \) by differentiating \( y \) with respect to time \( t \):
\[ \frac{dy}{dt} = \frac{d}{dt}(\sqrt{3}t - 5t^2 + t^3) \]
Calculating the derivative:
\[ V_y = \sqrt{3} - 10t + 3t^2 \]

### Step 2: Evaluate vertical velocity at \( t = 0 \)
To find the vertical component of the velocity at the moment of projection, we substitute \( t = 0 \):
\[ V_y(0) = \sqrt{3} - 10(0) + 3(0)^2 = \sqrt{3} \]

### Step 3: Differentiate the horizontal distance equation
The horizontal distance \( x \) is given by:
\[ x = t + 2t^2 - t^3 \]
We need to find the horizontal velocity \( V_x \) by differentiating \( x \) with respect to time \( t \):
\[ \frac{dx}{dt} = \frac{d}{dt}(t + 2t^2 - t^3) \]
Calculating the derivative:
\[ V_x = 1 + 4t - 3t^2 \]

### Step 4: Evaluate horizontal velocity at \( t = 0 \)
To find the horizontal component of the velocity at the moment of projection, we substitute \( t = 0 \):
\[ V_x(0) = 1 + 4(0) - 3(0)^2 = 1 \]

### Step 5: Calculate the angle of projection
The angle of projection \( \theta \) can be found using the relation:
\[ \tan(\theta) = \frac{V_y}{V_x} \]
Substituting the values we found:
\[ \tan(\theta) = \frac{\sqrt{3}}{1} = \sqrt{3} \]

### Step 6: Determine the angle \( \theta \)
The angle \( \theta \) for which \( \tan(\theta) = \sqrt{3} \) is:
\[ \theta = 60^\circ \]

### Final Answer
The angle of projection is \( 60^\circ \).

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