The equations of motion of a projectile are given by `x=36tm and 2y =96t-9.8t^(2)m`. The angle of projection is

Options

A. `sin^(-1)((4)/(5))`
B. `sin^(-1)((3)/(5))`
C. `sin^(-1)((4)/(3))`
D. `sin^(-1)((3)/(4))`

Explanation

To find the angle of projection for the given equations of motion of a projectile, we can follow these steps:

### Step 1: Identify the equations of motion
The equations of motion are given as:
- \( x = 36t \) (horizontal motion)
- \( 2y = 96t - 9.8t^2 \) (vertical motion)

### Step 2: Simplify the vertical motion equation
We can rewrite the vertical motion equation:
\[ y = 48t - 4.9t^2 \]

### Step 3: Find the horizontal component of the initial velocity
The horizontal component of the initial velocity \( u_x \) can be found by differentiating \( x \) with respect to time \( t \):
\[ u_x = \frac{dx}{dt} = \frac{d}{dt}(36t) = 36 \, \text{m/s} \]

### Step 4: Find the vertical component of the initial velocity
Similarly, we find the vertical component of the initial velocity \( u_y \) by differentiating \( y \) with respect to time \( t \):
\[ u_y = \frac{dy}{dt} = \frac{d}{dt}(48t - 4.9t^2) = 48 - 9.8t \]
At \( t = 0 \):
\[ u_y = 48 \, \text{m/s} \]

### Step 5: Relate the components to the angle of projection
The angle of projection \( \theta \) can be found using the relationship:
\[ \tan \theta = \frac{u_y}{u_x} \]
Substituting the values we found:
\[ \tan \theta = \frac{48}{36} = \frac{4}{3} \]

### Step 6: Calculate the angle \( \theta \)
To find \( \theta \), we can use the inverse tangent function:
\[ \theta = \tan^{-1}\left(\frac{4}{3}\right) \]

### Step 7: Determine sine of the angle
From the triangle formed by the components, we can find the hypotenuse \( h \) using the Pythagorean theorem:
\[ h = \sqrt{(4^2 + 3^2)} = \sqrt{16 + 9} = 5 \]
Now, we can find \( \sin \theta \):
\[ \sin \theta = \frac{\text{opposite}}{\text{hypotenuse}} = \frac{4}{5} \]

### Final Result
Thus, the angle of projection \( \theta \) can be expressed as:
\[ \theta = \sin^{-1}\left(\frac{4}{5}\right) \]