A girl standing on a stationary lift (open from above) throws a ball upwards with initial speed `50 m s^(-1)`. The time taken by the ball to return to her hands is (Take g = `10 m s^(-2)`)

Options

A. a. 5 s
B. b. 10 s
C. c. 15 s
D. d. 20 s

Explanation

To solve the problem step by step, we will analyze the motion of the ball thrown upwards by the girl in the stationary lift.

### Step 1: Identify the given values
- Initial speed of the ball, \( u = 50 \, \text{m/s} \)
- Acceleration due to gravity, \( g = 10 \, \text{m/s}^2 \) (acting downwards)

### Step 2: Understand the motion of the ball
When the girl throws the ball upwards, it will rise until its velocity becomes zero at the highest point. After reaching the highest point, the ball will fall back down to her hands.

### Step 3: Use the first equation of motion
We can use the first equation of motion to find the time taken to reach the highest point (let's call this time \( t \)):
\[ v = u + (-g)t \]
At the highest point, the final velocity \( v = 0 \). Therefore, we can rewrite the equation as:
\[ 0 = 50 - 10t \]

### Step 4: Solve for \( t \)
Rearranging the equation gives:
\[ 10t = 50 \]
\[ t = \frac{50}{10} = 5 \, \text{s} \]

### Step 5: Calculate the total time for the ball to return
The time taken to go up is equal to the time taken to come down. Therefore, the total time \( T \) for the ball to return to the girl's hands is:
\[ T = 2t = 2 \times 5 = 10 \, \text{s} \]

### Conclusion
The total time taken by the ball to return to the girl's hands is \( 10 \, \text{s} \).

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