What is the ratio of P.E. w.r.t. ground and K.E. at the top most point of the projectile motion:-

Options

A. `cos^(2)theta`
B. `sin^(2)theta`
C. `tan^(2)theta`
D. `cot^(2)theta`

Explanation

To find the ratio of potential energy (P.E.) with respect to the ground and kinetic energy (K.E.) at the topmost point of projectile motion, we can follow these steps:

### Step-by-Step Solution:

1. **Understanding the Projectile Motion:**
- In projectile motion, an object is launched at an angle θ with an initial velocity \( u \).
- The motion can be analyzed in two components: horizontal (x-axis) and vertical (y-axis).

2. **Identify the Kinetic Energy at the Topmost Point:**
- At the topmost point of the projectile's trajectory, the vertical component of the velocity becomes zero, and only the horizontal component remains.
- The horizontal component of the velocity is given by:
\[ v_x = u \cos \theta \]
- Therefore, the kinetic energy (K.E.) at the topmost point is:
\[ K.E. = \frac{1}{2} m v^2 = \frac{1}{2} m (u \cos \theta)^2 = \frac{1}{2} m u^2 \cos^2 \theta \quad \text{(Equation 1)} \]

3. **Calculate the Potential Energy at the Topmost Point:**
- The height (h) at the topmost point can be calculated using the formula for maximum height in projectile motion:
\[ h = \frac{u^2 \sin^2 \theta}{2g} \]
- The potential energy (P.E.) with respect to the ground at this height is:
\[ P.E. = mgh = mg \left(\frac{u^2 \sin^2 \theta}{2g}\right) = \frac{1}{2} m u^2 \sin^2 \theta \quad \text{(Equation 2)} \]

4. **Finding the Ratio of P.E. to K.E.:**
- Now, we need to find the ratio of potential energy to kinetic energy:
\[ \text{Ratio} = \frac{P.E.}{K.E.} = \frac{\frac{1}{2} m u^2 \sin^2 \theta}{\frac{1}{2} m u^2 \cos^2 \theta} \]
- The \( \frac{1}{2} m u^2 \) terms cancel out:
\[ \text{Ratio} = \frac{\sin^2 \theta}{\cos^2 \theta} = \tan^2 \theta \]

5. **Final Result:**
- Therefore, the ratio of potential energy with respect to the ground and kinetic energy at the topmost point of the projectile motion is:
\[ \tan^2 \theta \]