The vertical height of the projectile at the time is given by `y=4t-t^(2)` and the horizontal distance covered is given by `x=3t`. What is the angle of projection with the horizontal ?

Options

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

Answer

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

### Step 1: Understand the equations of motion
We are given the equations for vertical height (y) and horizontal distance (x) as functions of time (t):
- Vertical height: \( y = 4t - t^2 \)
- Horizontal distance: \( x = 3t \)

### Step 2: Differentiate to find velocities
To find the velocities in the x and y directions, we differentiate the equations with respect to time (t).

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

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

### Step 3: Find the initial vertical velocity
To find the initial vertical velocity, we evaluate \( v_y \) at \( t = 0 \):
\[ v_y(t=0) = 4 - 2(0) = 4 \, \text{m/s} \]

### Step 4: Write the velocity vector
Now we can express the velocity vector in terms of its components:
\[ \vec{v} = v_x \hat{i} + v_y \hat{j} = 3 \hat{i} + 4 \hat{j} \, \text{m/s} \]

### Step 5: Find the angle of projection
The angle of projection \( \theta \) can be found using the tangent function, which relates the opposite side (vertical component) to the adjacent side (horizontal component):
\[ \tan \theta = \frac{v_y}{v_x} = \frac{4}{3} \]

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

### Conclusion
Thus, the angle of projection with the horizontal is:
\[ \theta = \tan^{-1}\left(\frac{4}{3}\right) \]