A particle of mass.m. is projected with a velocity .u. at an angle . `theta`. with the horizontal. Work done by gravity during its descent from its highest point to a point which is at half the maximum height is

Options

A. 0
B. `(m u^(2) sin^(2)theta)/4`
C. `1/2m u^(2) sin^(2)theta`
D. `1/2m u^(2)cos^(2)theta`

Answer

To solve the problem of finding the work done by gravity during the descent of a particle from its highest point to a point that is at half the maximum height, we can follow these steps:

### Step-by-Step Solution:

1. **Identify the Maximum Height**:
The maximum height \( H \) reached by a projectile launched with an initial velocity \( u \) at an angle \( \theta \) is given by the formula:
\[ H = \frac{u^2 \sin^2 \theta}{2g} \]
where \( g \) is the acceleration due to gravity.

2. **Determine the Height for Work Done**:
The height \( h \) that the particle descends from the highest point to half of the maximum height is:
\[ h = \frac{H}{2} = \frac{1}{2} \left( \frac{u^2 \sin^2 \theta}{2g} \right) = \frac{u^2 \sin^2 \theta}{4g} \]

3. **Calculate the Work Done by Gravity**:
The work done by gravity \( W \) when the particle descends a height \( h \) is given by:
\[ W = mgh \]
Substituting the expression for \( h \) from step 2:
\[ W = mg \left( \frac{u^2 \sin^2 \theta}{4g} \right) \]

4. **Simplify the Expression**:
The \( g \) in the numerator and denominator cancels out:
\[ W = m \cdot \frac{u^2 \sin^2 \theta}{4} \]

5. **Final Result**:
Therefore, the work done by gravity during the descent from the highest point to half the maximum height is:
\[ W = \frac{m u^2 \sin^2 \theta}{4} \]