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 of finding the velocity of a packet dropped from a helicopter rising upwards with an initial velocity of 2 m/s after 2 seconds, we can use the first equation of motion. Here’s a step-by-step breakdown of the solution:

### Step 1: Identify the Initial Conditions
- The initial velocity (u) of the packet when it is dropped is equal to the velocity of the helicopter, which is **2 m/s** upwards. Therefore, we can write:
\[ u = +2 \, \text{m/s} \]
- The acceleration (a) acting on the packet is due to gravity, which acts downwards. We take downward as negative, so:
\[ a = -10 \, \text{m/s}^2 \]

### Step 2: Use the First Equation of Motion
The first equation of motion states:
\[ v = u + at \]
where:
- \( v \) is the final velocity,
- \( u \) is the initial velocity,
- \( a \) is the acceleration,
- \( t \) is the time.

### Step 3: Substitute the Known Values
We need to find the velocity after \( t = 2 \) seconds. Substituting the values into the equation:
\[ v = 2 \, \text{m/s} + (-10 \, \text{m/s}^2) \times (2 \, \text{s}) \]

### Step 4: Calculate the Result
Now, calculate the acceleration term:
\[ v = 2 \, \text{m/s} - 20 \, \text{m/s} \]
\[ v = -18 \, \text{m/s} \]

### Step 5: Interpret the Result
The negative sign indicates that the velocity is directed downwards. Thus, the magnitude of the velocity is:
\[ |v| = 18 \, \text{m/s} \text{ (downwards)} \]

### Final Answer
The velocity of the packet after 2 seconds is **18 m/s downwards**.

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