A man in a moving bus drops a bag from a height 2 m. The bag lands on the ground with a velocity 10 m/s. The velocity of the bus at the moment he dropped the bag was:

Options

A. `10 m//s`
B. `2sqrt(10) m//s`
C. `2sqrt(15) m//s`
D. `5 m//s`

Explanation

To solve the problem step by step, we will analyze the motion of the bag dropped from the bus and apply the principles of kinematics.

### Step 1: Understand the scenario
A man drops a bag from a height of 2 meters while the bus is moving horizontally. The bag will have the same horizontal velocity as the bus at the moment it is dropped.

### Step 2: Identify the known values
- Height (h) = 2 m
- Final velocity of the bag when it hits the ground (v) = 10 m/s
- Acceleration due to gravity (g) = 10 m/s² (assuming standard value)

### Step 3: Analyze vertical motion
The bag is dropped from a height of 2 m, and we need to find the vertical component of the velocity (v_y) when it hits the ground. We can use the third equation of motion for vertical motion:
\[ v_y^2 = u_y^2 + 2gh \]
Where:
- \( u_y \) = initial vertical velocity = 0 (since the bag is dropped)
- \( g \) = acceleration due to gravity = 10 m/s²
- \( h \) = height = 2 m

Substituting the values:
\[ v_y^2 = 0 + 2 \cdot 10 \cdot 2 \]
\[ v_y^2 = 40 \]
\[ v_y = \sqrt{40} = 2\sqrt{10} \, \text{m/s} \]

### Step 4: Relate horizontal and vertical velocities
At the moment the bag hits the ground, it has both horizontal and vertical components of velocity. The total velocity (v) can be found using the Pythagorean theorem:
\[ v^2 = u^2 + v_y^2 \]
Where:
- \( u \) = horizontal velocity of the bus (which we need to find)
- \( v_y \) = vertical velocity we just calculated
- \( v \) = total velocity when it hits the ground = 10 m/s

Substituting the known values:
\[ 10^2 = u^2 + (2\sqrt{10})^2 \]
\[ 100 = u^2 + 40 \]

### Step 5: Solve for the horizontal velocity (u)
Rearranging the equation:
\[ u^2 = 100 - 40 \]
\[ u^2 = 60 \]
\[ u = \sqrt{60} = \sqrt{4 \cdot 15} = 2\sqrt{15} \, \text{m/s} \]

### Final Answer
The velocity of the bus at the moment he dropped the bag was \( 2\sqrt{15} \, \text{m/s} \).

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