A body is projected at an angle of `45^(@)` with horizontal with velocity of `40sqrt(2)m//s`. Find out
(i) Maximum height attained by body
(ii) Time of flight
(iii) Horizontal range
(iv) Velocity at maximum height
(v) The ratio of Kinetic energy to potential energy at highest point
(vi) The part equation of projectile, assuming point of projection as origin and consider x and y are horizontal and vertical distance in meter
(vii) The horizontal distance covered by body in 2 second
(viii) The vertical distance covered by body in 2 second

Answer

To solve the problem step by step, we will break down each part of the question systematically.

### Given Data:
- Initial velocity, \( u = 40\sqrt{2} \, \text{m/s} \)
- Angle of projection, \( \theta = 45^\circ \)
- Acceleration due to gravity, \( g = 10 \, \text{m/s}^2 \)

### Step 1: Maximum Height Attained by the Body
The formula for maximum height \( h_{max} \) is given by:
\[ h_{max} = \frac{u^2 \sin^2 \theta}{2g} \]
Substituting the values:
\[ h_{max} = \frac{(40\sqrt{2})^2 \cdot \left(\frac{1}{\sqrt{2}}\right)^2}{2 \cdot 10} \]
Calculating:
\[ = \frac{3200 \cdot \frac{1}{2}}{20} = \frac{1600}{20} = 80 \, \text{m} \]

### Step 2: Time of Flight
The formula for time of flight \( T \) is:
\[ T = \frac{2u \sin \theta}{g} \]
Substituting the values:
\[ T = \frac{2 \cdot 40\sqrt{2} \cdot \frac{1}{\sqrt{2}}}{10} \]
Calculating:
\[ = \frac{80}{10} = 8 \, \text{s} \]

### Step 3: Horizontal Range
The formula for horizontal range \( R \) is:
\[ R = \frac{u^2 \sin 2\theta}{g} \]
Substituting the values:
\[ R = \frac{(40\sqrt{2})^2 \cdot \sin(90^\circ)}{10} \]
Calculating:
\[ = \frac{3200 \cdot 1}{10} = 320 \, \text{m} \]

### Step 4: Velocity at Maximum Height
At maximum height, the vertical component of velocity is zero. The horizontal component remains:
\[ v_{max\ height} = u \cos \theta \]
Calculating:
\[ = 40\sqrt{2} \cdot \frac{1}{\sqrt{2}} = 40 \, \text{m/s} \]

### Step 5: Ratio of Kinetic Energy to Potential Energy at Highest Point
Kinetic Energy \( KE \) at maximum height:
\[ KE = \frac{1}{2} m v^2 = \frac{1}{2} m (40)^2 = 800m \]
Potential Energy \( PE \) at maximum height:
\[ PE = mgh_{max} = mg \cdot 80 = 800m \]
The ratio \( \frac{KE}{PE} \):
\[ \frac{KE}{PE} = \frac{800m}{800m} = 1:1 \]

### Step 6: Parametric Equations of Projectile
The parametric equations for the projectile motion are:
\[ x(t) = u \cos \theta \cdot t = 40t \]
\[ y(t) = u \sin \theta \cdot t - \frac{1}{2} g t^2 = 40t - 5t^2 \]

### Step 7: Horizontal Distance Covered in 2 Seconds
Using the equation for horizontal distance:
\[ x(2) = 40 \cdot 2 = 80 \, \text{m} \]

### Step 8: Vertical Distance Covered in 2 Seconds
Using the equation for vertical distance:
\[ y(2) = 40 \cdot 2 - 5 \cdot (2)^2 = 80 - 20 = 60 \, \text{m} \]

### Summary of Results:
1. Maximum height: \( 80 \, \text{m} \)
2. Time of flight: \( 8 \, \text{s} \)
3. Horizontal range: \( 320 \, \text{m} \)
4. Velocity at maximum height: \( 40 \, \text{m/s} \)
5. Ratio of KE to PE at highest point: \( 1:1 \)
6. Parametric equations: \( x(t) = 40t, \, y(t) = 40t - 5t^2 \)
7. Horizontal distance in 2 seconds: \( 80 \, \text{m} \)
8. Vertical distance in 2 seconds: \( 60 \, \text{m} \)