A packet is dropped from a helicopter rising up with velocity `2 ms^(-1)`. The velocity of the packet after 2 seconds will be

Options

A. 18 m/s
B. 10 m/s
C. 8 m/s
D. 2 m/s

Answer

To solve the problem, we need to determine the velocity of the packet after it is dropped from the helicopter. Here’s a step-by-step solution:

### Step 1: Identify the initial conditions
- The helicopter is rising with a velocity of \( u = 2 \, \text{m/s} \) (upward).
- The packet is dropped from the helicopter, so its initial velocity is also \( u = 2 \, \text{m/s} \) (upward).
- The acceleration due to gravity is \( g = 10 \, \text{m/s}^2 \) (downward).

### Step 2: Set the direction
- We will take the upward direction as positive. Hence, the acceleration due to gravity will be negative: \( a = -g = -10 \, \text{m/s}^2 \).

### Step 3: Use the kinematic equation
We will use the kinematic equation:
\[ v = u + at \]
Where:
- \( v \) is the final velocity,
- \( u \) is the initial velocity,
- \( a \) is the acceleration,
- \( t \) is the time.

### Step 4: Substitute the values
Substituting the known values into the equation:
\[ v = 2 \, \text{m/s} + (-10 \, \text{m/s}^2)(2 \, \text{s}) \]
\[ v = 2 \, \text{m/s} - 20 \, \text{m/s} \]
\[ v = -18 \, \text{m/s} \]

### Step 5: Interpret the result
The negative sign indicates that the velocity is directed downward. Therefore, the magnitude of the velocity is \( 18 \, \text{m/s} \) and the direction is downward.

### Final Answer
The velocity of the packet after 2 seconds is \( 18 \, \text{m/s} \) downward.

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