A projectile is projected with a speed of 40m/s at a angle `theta` with horizontal such that `tantheta=(3)/(4)`. After 2 sec, the projectile is moving with speed v at an angle `alpha` with horizontal then, `(g=10m//s^(2))`

Options

A. `tanalpha=(1)/(8)`
B. `cot alpha=(1)/(8)`
C. `v=32.25m//s`
D. `v=32m//s`

Answer

To solve the problem step by step, we will break down the projectile motion into its components and apply the relevant equations of motion.

### Step 1: Determine the initial velocity components
Given:
- Initial speed \( u = 40 \, \text{m/s} \)
- \( \tan \theta = \frac{3}{4} \)

Using the tangent function, we can find the components of the initial velocity:
\[ \sin \theta = \frac{3}{5} \quad \text{and} \quad \cos \theta = \frac{4}{5} \]
Thus, the horizontal and vertical components of the initial velocity are:
\[ u_x = u \cos \theta = 40 \cdot \frac{4}{5} = 32 \, \text{m/s} \]
\[ u_y = u \sin \theta = 40 \cdot \frac{3}{5} = 24 \, \text{m/s} \]

### Step 2: Analyze the vertical motion after 2 seconds
The vertical motion is influenced by gravity. The vertical component of the velocity after time \( t \) can be calculated using the equation:
\[ v_y = u_y - g t \]
Substituting the known values:
\[ v_y = 24 - 10 \cdot 2 = 24 - 20 = 4 \, \text{m/s} \]

### Step 3: Determine the horizontal component of velocity
Since there are no horizontal forces acting on the projectile, the horizontal component of the velocity remains constant:
\[ v_x = u_x = 32 \, \text{m/s} \]

### Step 4: Calculate the resultant velocity \( v \)
The resultant velocity can be found using the Pythagorean theorem:
\[ v = \sqrt{v_x^2 + v_y^2} = \sqrt{32^2 + 4^2} = \sqrt{1024 + 16} = \sqrt{1040} \approx 32.25 \, \text{m/s} \]

### Step 5: Determine the angle \( \alpha \)
To find the angle \( \alpha \) with respect to the horizontal, we use the tangent function:
\[ \tan \alpha = \frac{v_y}{v_x} = \frac{4}{32} = \frac{1}{8} \]

### Summary of Results
- The speed \( v \) after 2 seconds is approximately \( 32.25 \, \text{m/s} \).
- The angle \( \alpha \) satisfies \( \tan \alpha = \frac{1}{8} \).