A balloon is at a height of 40 m and is ascending with a velocity of `10 ms^(-1)`. A bag of 5 kg weight is dropped from it, when will the bag reach surface of earth ? `(g = 10 m//s^(2))`

Options

A. 2 sec
B. 4 sec
C. 6 sec
D. `2 sqrt(2) sec`

Answer

To solve the problem of when the bag will reach the surface of the Earth after being dropped from a balloon, we can use the equations of motion. Here’s a step-by-step solution:

### Step 1: Identify the initial conditions
- The height from which the bag is dropped, \( S = -40 \, \text{m} \) (negative because it is falling down).
- The initial velocity of the bag when dropped, \( U = 10 \, \text{m/s} \) (upward, so we consider it positive).
- The acceleration due to gravity, \( g = 10 \, \text{m/s}^2 \) (downward, so we consider it negative).

### Step 2: Use the equation of motion
We will use the second equation of motion:
\[ S = Ut + \frac{1}{2} a t^2 \]
Substituting the values:
\[ -40 = 10t - \frac{1}{2} \cdot 10 \cdot t^2 \]
This simplifies to:
\[ -40 = 10t - 5t^2 \]

### Step 3: Rearrange the equation
Rearranging the equation gives:
\[ 5t^2 - 10t - 40 = 0 \]

### Step 4: Simplify the equation
Dividing the entire equation by 5:
\[ t^2 - 2t - 8 = 0 \]

### Step 5: Factor the quadratic equation
We can factor the quadratic equation:
\[ (t - 4)(t + 2) = 0 \]

### Step 6: Solve for \( t \)
Setting each factor to zero gives:
1. \( t - 4 = 0 \) → \( t = 4 \, \text{s} \)
2. \( t + 2 = 0 \) → \( t = -2 \, \text{s} \) (not physically meaningful)

Thus, the time taken for the bag to reach the surface of the Earth is:
\[ t = 4 \, \text{s} \]

### Final Answer
The bag will reach the surface of the Earth in **4 seconds**.

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