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, we need to find the velocity of the bus at the moment the bag was dropped. Here’s a step-by-step breakdown of the solution:

### Step 1: Understand the motion of the bag
When the man drops the bag from the bus, the bag has the same horizontal velocity as the bus at the moment of release. The bag will then fall under the influence of gravity and will also continue to move horizontally with the velocity of the bus.

### Step 2: Analyze the vertical motion
The bag is dropped from a height of 2 meters. We can use the kinematic equation for vertical motion to find the vertical component of the velocity of the bag just before it hits the ground. The equation is:
\[ v_y^2 = u_y^2 + 2gh \]
Where:
- \( v_y \) = final vertical velocity
- \( u_y \) = initial vertical velocity (0 m/s, since it is dropped)
- \( g \) = acceleration due to gravity (approximately \( 10 \, \text{m/s}^2 \))
- \( h \) = height (2 m)

### Step 3: Calculate the vertical velocity
Substituting the values into the equation:
\[ v_y^2 = 0 + 2 \times 10 \times 2 \]
\[ v_y^2 = 40 \]
\[ v_y = \sqrt{40} = 2\sqrt{10} \, \text{m/s} \]

### Step 4: Relate the velocities
The final velocity of the bag when it hits the ground is given as 10 m/s. This velocity is the resultant of the horizontal and vertical components of the velocity:
\[ v^2 = v_x^2 + v_y^2 \]
Where:
- \( v \) = total velocity (10 m/s)
- \( v_x \) = horizontal velocity (which is the velocity of the bus)
- \( v_y \) = vertical velocity (calculated in Step 3)

### Step 5: Substitute and solve for \( v_x \)
Substituting the known values:
\[ 10^2 = v_x^2 + (2\sqrt{10})^2 \]
\[ 100 = v_x^2 + 40 \]
\[ v_x^2 = 100 - 40 \]
\[ v_x^2 = 60 \]
\[ v_x = \sqrt{60} = \sqrt{4 \times 15} = 2\sqrt{15} \, \text{m/s} \]

### Step 6: Conclusion
The velocity of the bus at the moment he dropped the bag is:
\[ v_x = 2\sqrt{15} \, \text{m/s} \]