A body of mass m is thrown upwards at an angle `theta` with the horizontal with velocity v. While rising up the velocity of the mass after t second will be

Options

A. `sqrt((v cos theta^(2)) + (v sin theta)^(2))`
B. `sqrt((v cos theta - v sin theta)^(2) - g t)`
C. `sqrt(v^(2) + g^(2)t^(2) - (2b sin theta) g t)`
D. `sqrt(v^(2) + g^(2)t^(2) -(2v cos theta) g t)`

Explanation

To find the velocity of a body of mass \( m \) thrown upwards at an angle \( \theta \) with an initial velocity \( v \) after \( t \) seconds, we can break down the problem into the horizontal and vertical components of the motion.

### Step-by-Step Solution:

1. **Identify the Components of Initial Velocity**:
- The initial velocity \( v \) can be resolved into two components:
- Horizontal component: \( v_x = v \cos \theta \)
- Vertical component: \( v_y = v \sin \theta \)

2. **Determine the Horizontal Velocity**:
- The horizontal motion has no acceleration (assuming air resistance is negligible), so the horizontal component of velocity remains constant:
\[ v_x = v \cos \theta \]

3. **Determine the Vertical Velocity After \( t \) Seconds**:
- The vertical motion is affected by gravity. The acceleration due to gravity \( g \) acts downward, which we take as negative:
- The vertical velocity after \( t \) seconds can be calculated using the kinematic equation:
\[ v_y = u_y + at \]
where \( u_y = v \sin \theta \) (initial vertical velocity), \( a = -g \) (acceleration due to gravity), and \( t \) is the time.
- Substituting the values:
\[ v_y = v \sin \theta - g t \]

4. **Calculate the Magnitude of the Resultant Velocity**:
- The resultant velocity \( v \) can be found using the Pythagorean theorem since the horizontal and vertical components are perpendicular to each other:
\[ v = \sqrt{v_x^2 + v_y^2} \]
- Substituting the expressions for \( v_x \) and \( v_y \):
\[ v = \sqrt{(v \cos \theta)^2 + (v \sin \theta - g t)^2} \]

5. **Simplify the Expression**:
- Expanding the squares:
\[ v = \sqrt{v^2 \cos^2 \theta + (v \sin \theta - g t)^2} \]
- This expands to:
\[ v = \sqrt{v^2 \cos^2 \theta + (v^2 \sin^2 \theta - 2v \sin \theta \cdot g t + g^2 t^2)} \]
- Combine \( v^2 \cos^2 \theta \) and \( v^2 \sin^2 \theta \):
\[ v = \sqrt{v^2 (\cos^2 \theta + \sin^2 \theta) + g^2 t^2 - 2v \sin \theta \cdot g t} \]
- Using the identity \( \cos^2 \theta + \sin^2 \theta = 1 \):
\[ v = \sqrt{v^2 + g^2 t^2 - 2v \sin \theta \cdot g t} \]

### Final Result:
The final velocity of the mass after \( t \) seconds is given by:
\[ v = \sqrt{v^2 + g^2 t^2 - 2v g t \sin \theta} \]