A body of mass m is projected from ground with speed u at an angle `theta` with horizontal. The power delivered by gravity to it at half of maximum height from ground is

Options

A. `(mgu cos theta)/(sqrt2)`
B. `(mgu sin theta)/(sqrt2)`
C. `(mgu cos (90+theta))/(sqrt2)`
D. Both (2) & (3)

Answer

To find the power delivered by gravity to a body of mass \( m \) projected from the ground with speed \( u \) at an angle \( \theta \) with the horizontal, we will follow these steps:

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

### Step 2: Calculate Half of the Maximum Height
To find half of the maximum height, we divide \( H \) by 2:
\[ h = \frac{H}{2} = \frac{u^2 \sin^2 \theta}{4g} \]

### Step 3: Calculate the Vertical Component of Initial Velocity
The vertical component of the initial velocity \( u_y \) is given by:
\[ u_y = u \sin \theta \]

### Step 4: Use the Third Equation of Motion to Find Final Vertical Velocity
Using the third equation of motion, we can find the vertical velocity \( v_y \) at height \( h \):
\[ v_y^2 = u_y^2 - 2gh \]
Substituting \( u_y \) and \( h \):
\[ v_y^2 = (u \sin \theta)^2 - 2g \left(\frac{u^2 \sin^2 \theta}{4g}\right) \]
This simplifies to:
\[ v_y^2 = u^2 \sin^2 \theta - \frac{u^2 \sin^2 \theta}{2} = \frac{u^2 \sin^2 \theta}{2} \]
Taking the square root gives:
\[ v_y = \frac{u \sin \theta}{\sqrt{2}} \]

### Step 5: Calculate the Power Delivered by Gravity
The power \( P \) delivered by gravity can be calculated using:
\[ P = F \cdot v_y \]
where \( F = mg \) (the force due to gravity). Therefore:
\[ P = mg \cdot \frac{u \sin \theta}{\sqrt{2}} \]

### Final Expression for Power
Thus, the power delivered by gravity at half of the maximum height is:
\[ P = \frac{mgu \sin \theta}{\sqrt{2}} \]