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
Horizontal range of the ground is …………`m`

Options

A. 60
B. 50
C. 80
D. 70

Answer

To find the horizontal range of the projectile given the velocity at a height of 15 m, we can follow these steps:

### Step-by-Step Solution:

1. **Identify the Given Information:**
- The velocity of the projectile at a height of 15 m is given as \( \vec{v} = 20 \hat{i} + 10 \hat{j} \) m/s.
- Here, \( \hat{i} \) represents the horizontal component and \( \hat{j} \) represents the vertical component.

2. **Extract the Components of Velocity:**
- The horizontal component of velocity, \( v_x = 20 \) m/s.
- The vertical component of velocity, \( v_y = 10 \) m/s.

3. **Use the Kinematic Equation to Find Initial Vertical Velocity:**
- We can use the kinematic equation:
\[ v_y^2 = u_y^2 + 2a_y s \]
- Here, \( v_y \) is the final vertical velocity (10 m/s), \( u_y \) is the initial vertical velocity, \( a_y \) is the acceleration due to gravity (-10 m/s², acting downwards), and \( s \) is the height (15 m).
- Rearranging gives:
\[ u_y^2 = v_y^2 - 2a_y s \]
- Substituting the known values:
\[ u_y^2 = 10^2 - 2 \cdot (-10) \cdot 15 \]
\[ u_y^2 = 100 + 300 = 400 \]
\[ u_y = \sqrt{400} = 20 \text{ m/s} \]

4. **Determine the Horizontal Component of Initial Velocity:**
- From the given data, we already know \( u_x = 20 \) m/s.

5. **Calculate the Time of Flight to Reach the Ground:**
- The time to reach the maximum height can be calculated using:
\[ t = \frac{u_y}{g} \]
- For \( u_y = 20 \) m/s and \( g = 10 \) m/s²:
\[ t = \frac{20}{10} = 2 \text{ seconds} \]
- The total time of flight will be double this (up and down):
\[ T = 2t = 2 \cdot 2 = 4 \text{ seconds} \]

6. **Calculate the Horizontal Range:**
- The horizontal range \( R \) can be calculated using:
\[ R = u_x \cdot T \]
- Substituting the values:
\[ R = 20 \cdot 4 = 80 \text{ meters} \]

### Final Answer:
The horizontal range of the projectile is **80 meters**.

---