At a height of 45 m from ground velocity of a projectile is,
`v = (30hati + 40hatj) m//s`
Find initial velocity, time of flight, maximum height and horizontal range of this projectile. Here `hati` and` hatj` are the unit vectors in horizontal and vertical directions.

Explanation

To solve the problem, we will follow these steps:

### Step 1: Identify Given Information
- The height (h) at which the velocity is given: \( h = 45 \, \text{m} \)
- The velocity at this height:
\[ \mathbf{v} = 30 \hat{i} + 40 \hat{j} \, \text{m/s} \]
- Horizontal component (\( v_x \)): \( 30 \, \text{m/s} \)
- Vertical component (\( v_y \)): \( 40 \, \text{m/s} \)

### Step 2: Find the Initial Vertical Velocity (\( u_y \))
Using the equation of motion in the vertical direction:
\[ v_y^2 = u_y^2 + 2gh \]
Rearranging gives:
\[ u_y^2 = v_y^2 - 2gh \]
Substituting the known values (\( g = 10 \, \text{m/s}^2 \)):
\[ u_y^2 = 40^2 - 2 \cdot 10 \cdot 45 \]
Calculating:
\[ u_y^2 = 1600 - 900 = 700 \]
Thus,
\[ u_y = \sqrt{700} = 10\sqrt{7} \approx 26.46 \, \text{m/s} \]

### Step 3: Find the Initial Horizontal Velocity (\( u_x \))
Since the horizontal component of velocity remains constant:
\[ u_x = v_x = 30 \, \text{m/s} \]

### Step 4: Calculate the Magnitude of the Initial Velocity (\( u \))
Using Pythagoras' theorem:
\[ u = \sqrt{u_x^2 + u_y^2} \]
Substituting the values:
\[ u = \sqrt{30^2 + (10\sqrt{7})^2} = \sqrt{900 + 700} = \sqrt{1600} = 40 \, \text{m/s} \]

### Step 5: Calculate the Angle of Projection (\( \theta \))
Using the tangent function:
\[ \tan \theta = \frac{u_y}{u_x} = \frac{10\sqrt{7}}{30} = \frac{\sqrt{7}}{3} \]
Thus,
\[ \theta = \tan^{-1}\left(\frac{\sqrt{7}}{3}\right) \]

### Step 6: Calculate Time of Flight (\( T \))
Using the formula for time of flight:
\[ T = \frac{2u_y}{g} \]
Substituting the values:
\[ T = \frac{2 \cdot 10\sqrt{7}}{10} = 2\sqrt{7} \approx 5.29 \, \text{s} \]

### Step 7: Calculate Maximum Height (\( H \))
Using the formula for maximum height:
\[ H = \frac{u_y^2}{2g} \]
Substituting the values:
\[ H = \frac{(10\sqrt{7})^2}{2 \cdot 10} = \frac{700}{20} = 35 \, \text{m} \]

### Step 8: Calculate Horizontal Range (\( R \))
Using the formula for horizontal range:
\[ R = u_x \cdot T \]
Substituting the values:
\[ R = 30 \cdot 2\sqrt{7} \approx 30 \cdot 5.29 \approx 158.7 \, \text{m} \]

### Summary of Results
- Initial Velocity (\( u \)): \( 40 \, \text{m/s} \)
- Time of Flight (\( T \)): \( 5.29 \, \text{s} \)
- Maximum Height (\( H \)): \( 35 \, \text{m} \)
- Horizontal Range (\( R \)): \( 158.7 \, \text{m} \)