A particle of mass m is projected with speed u at angle `theta` with horizontal from ground. The work done by gravity on it during its upward motion is

Options

A. `(-m u^(2)sin^(2)theta)/(2)`
B. `(m u^(2)cos^(2)theta)/(2)`
C. `(m u^(2)sin^(2)theta)/(2)`
D. Zero

Answer

To solve the problem of finding the work done by gravity on a particle of mass \( m \) projected with speed \( u \) at an angle \( \theta \) with the horizontal during its upward motion, we can follow these steps:

### Step 1: Determine the maximum height reached by the particle
The maximum height \( h_{\text{max}} \) reached by the projectile can be calculated using the formula:
\[ h_{\text{max}} = \frac{u^2 \sin^2 \theta}{2g} \]
where \( g \) is the acceleration due to gravity.

### Step 2: Identify the force acting on the particle
The force acting on the particle due to gravity is given by:
\[ F = mg \]
This force acts downward throughout the motion.

### Step 3: Calculate the work done by gravity
The work done by a force is given by the formula:
\[ W = F \cdot d \cdot \cos(\phi) \]
where \( F \) is the force, \( d \) is the displacement, and \( \phi \) is the angle between the force and the displacement vector.

In this case:
- The force \( F = mg \)
- The displacement \( d = h_{\text{max}} \)
- The angle \( \phi = 180^\circ \) (since gravity acts downward while the displacement during the upward motion is upward)

Thus, we have:
\[ W = mg \cdot h_{\text{max}} \cdot \cos(180^\circ) \]
Since \( \cos(180^\circ) = -1 \), we can rewrite the equation as:
\[ W = -mg \cdot h_{\text{max}} \]

### Step 4: Substitute the expression for maximum height
Now substituting the expression for \( h_{\text{max}} \) into the work done equation:
\[ W = -mg \cdot \left( \frac{u^2 \sin^2 \theta}{2g} \right) \]

### Step 5: Simplify the expression
Now, we can simplify the expression:
\[ W = -\frac{m u^2 \sin^2 \theta}{2} \]

### Final Answer
Thus, the work done by gravity on the particle during its upward motion is:
\[ W = -\frac{m u^2 \sin^2 \theta}{2} \]