Velocity of a projectile at height `15 m` from ground is `v=(20hati+10hatj)m//s`. Here `hati` is in horizontal direction and `hatj` is vertically upwards. Then
Speed with which particle is projected from ground is……… m/s

Options

A. 30
B. `20sqrt(2)`
C. `sqrt(20)`
D. `3sqrt(40)`

Answer

To find the speed with which the particle is projected from the ground, we can follow these steps:

### Step 1: Understand the given information
We are given the velocity of the projectile at a height of 15 m:
\[ \mathbf{v} = 20 \hat{i} + 10 \hat{j} \, \text{m/s} \]
Here, \( \hat{i} \) represents the horizontal direction and \( \hat{j} \) represents the vertical direction.

### Step 2: Identify the components of velocity
From the given velocity:
- The horizontal component of velocity, \( v_x = 20 \, \text{m/s} \)
- The vertical component of velocity, \( v_y = 10 \, \text{m/s} \)

### Step 3: Use the equations of motion
In the vertical direction, the acceleration is due to gravity, which is \( a_y = -g = -10 \, \text{m/s}^2 \).

Using the third equation of motion:
\[ v_y^2 = u_y^2 + 2a_y s \]
Where:
- \( v_y = 10 \, \text{m/s} \) (vertical component of velocity at height)
- \( u_y \) is the initial vertical component of velocity (which we need to find)
- \( a_y = -10 \, \text{m/s}^2 \) (acceleration due to gravity)
- \( s = 15 \, \text{m} \) (height)

### Step 4: Substitute the values into the equation
Substituting the known values into the equation:
\[ (10)^2 = u_y^2 + 2(-10)(15) \]
\[ 100 = u_y^2 - 300 \]
\[ u_y^2 = 100 + 300 = 400 \]
\[ u_y = \sqrt{400} = 20 \, \text{m/s} \]

### Step 5: Find the horizontal component of initial velocity
Since there is no horizontal acceleration, the horizontal component of the initial velocity \( u_x \) is the same as \( v_x \):
\[ u_x = v_x = 20 \, \text{m/s} \]

### Step 6: Calculate the initial speed of projection
Now we can find the initial speed \( u \) using the Pythagorean theorem:
\[ u = \sqrt{u_x^2 + u_y^2} \]
Substituting the values:
\[ u = \sqrt{(20)^2 + (20)^2} = \sqrt{400 + 400} = \sqrt{800} = 20\sqrt{2} \, \text{m/s} \]

### Final Answer
The speed with which the particle is projected from the ground is:
\[ \boxed{20\sqrt{2} \, \text{m/s}} \]