Trajectory of particle in a projectile motion is given as `y=x- x^(2)/80`. Here, x and y are in meters. For this projectile motion, match the following with `g=10 m//s^(2)`.
`{:(,"Column-I",,"Column-II"),((A),"Angle of projection (in degrees)",(P),20),((B),"Angle of velocity with horizontal after 4s (in degrees)",(Q),80),((C),"Maximum height (in meters)",(R),45),((D),"Horizontal range (in meters)",(S),30),(,,(T),60):}`

Answer

To solve the problem, we will analyze the given trajectory equation of the projectile motion and extract the required information step by step.

### Step 1: Identify the trajectory equation
The trajectory of the particle is given as:
\[ y = x - \frac{x^2}{80} \]

### Step 2: Compare with the standard trajectory equation
The standard form of the trajectory for projectile motion is:
\[ y = x \tan \theta - \frac{g}{2u^2 \cos^2 \theta} x^2 \]
where \( \theta \) is the angle of projection, \( g \) is the acceleration due to gravity, and \( u \) is the initial velocity.

### Step 3: Determine the angle of projection
From the given equation, we can see that:
- The coefficient of \( x \) is 1, which corresponds to \( x \tan \theta \).
- The coefficient of \( x^2 \) is \(-\frac{1}{80}\), which corresponds to \(-\frac{g}{2u^2 \cos^2 \theta}\).

Since \( g = 10 \, \text{m/s}^2 \), we can set up the following equations:
1. \( \tan \theta = 1 \) (from the coefficient of \( x \))
2. \( \frac{g}{2u^2 \cos^2 \theta} = \frac{1}{80} \)

From \( \tan \theta = 1 \), we have:
\[ \theta = 45^\circ \]

### Step 4: Calculate the horizontal range
The horizontal range \( R \) for projectile motion is given by:
\[ R = \frac{u^2 \sin 2\theta}{g} \]

Since \( \theta = 45^\circ \), we have:
\[ \sin 2\theta = \sin 90^\circ = 1 \]

Thus,
\[ R = \frac{u^2}{g} \]

From the comparison of coefficients, we know that:
\[ R = 80 \, \text{m} \]

### Step 5: Calculate the maximum height
The maximum height \( H \) for projectile motion is given by:
\[ H = \frac{R}{4} \]

Substituting \( R = 80 \):
\[ H = \frac{80}{4} = 20 \, \text{m} \]

### Step 6: Calculate the angle of velocity with horizontal after 4 seconds
The time of flight \( T \) for projectile motion is given by:
\[ T = \frac{2u \sin \theta}{g} \]

Since \( \theta = 45^\circ \):
\[ T = \frac{2u \cdot \frac{1}{\sqrt{2}}}{g} = \frac{u\sqrt{2}}{5} \]

Given that the total time of flight is \( 8 \, \text{s} \) (since it takes 4 seconds to reach the maximum height), we can find \( u \):
\[ 8 = \frac{u\sqrt{2}}{5} \implies u = 20\sqrt{2} \]

At \( t = 4 \, \text{s} \), the angle of velocity with horizontal is still \( 45^\circ \) because the projectile is symmetric in its motion.

### Final Answers
- (A) Angle of projection: \( 45^\circ \) (matches with R)
- (B) Angle of velocity with horizontal after 4s: \( 45^\circ \) (matches with R)
- (C) Maximum height: \( 20 \, \text{m} \) (matches with P)
- (D) Horizontal range: \( 80 \, \text{m} \) (matches with S)