A particle of mass `m` is projected at an angle `alpha` to the horizontal with an initial velocity `u`. The work done by gravity during the time it reaches its highest point is

Options

A. (a) `u^2sin^2alpha`
B. (b) `(m u^2cos^2alpha)/(2)`
C. (c) `(m u^2sin^2alpha)/(2)`
D. (d) `-(m u^2sin^2alpha)/(2)`

Explanation

To solve the problem of finding the work done by gravity during the time a particle of mass \( m \) reaches its highest point when projected at an angle \( \alpha \) with an initial velocity \( u \), we can follow these steps:

### Step 1: Understand the concept of work done by gravity
The work done by gravity when an object moves vertically is given by the formula:
\[ W = -mgh \]
where \( h \) is the height the object rises to, and the negative sign indicates that gravity does work against the direction of motion.

### Step 2: Determine the maximum height \( h \)
In projectile motion, the maximum height \( h \) reached by the particle can be calculated using the formula:
\[ h = \frac{u^2 \sin^2 \alpha}{2g} \]
This formula arises from the kinematic equations of motion under uniform acceleration due to gravity.

### Step 3: Substitute the height into the work done formula
Now, substituting the expression for \( h \) into the work done by gravity:
\[ W = -mg \left( \frac{u^2 \sin^2 \alpha}{2g} \right) \]

### Step 4: Simplify the expression
The \( g \) in the numerator and denominator cancels out:
\[ W = -m \frac{u^2 \sin^2 \alpha}{2} \]

### Step 5: Write the final expression for work done
Thus, the work done by gravity during the time it reaches its highest point is:
\[ W = -\frac{mu^2 \sin^2 \alpha}{2} \]

### Final Answer:
The work done by gravity is:
\[ W = -\frac{mu^2 \sin^2 \alpha}{2} \]

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