A player throws a ball upwards with an initial speed of `30 m s^(-1)`. How long does the ball take to return to the player's hands? ( Take g = 10 `m s^(-2)`)

Options

A. a. 3 s
B. b. 6 s
C. c. 9 s
D. d. 12 s

Explanation

To solve the problem of how long it takes for a ball thrown upwards at an initial speed of 30 m/s to return to the player's hands, we can follow these steps:

### Step 1: Identify the given values
- Initial velocity (u) = 30 m/s (upwards)
- Acceleration due to gravity (g) = -10 m/s² (negative because it acts downwards)

### Step 2: Determine the time taken to reach the highest point
At the highest point of its trajectory, the final velocity (V) of the ball will be 0 m/s. We can use the first equation of motion:
\[ V = u + g \cdot t \]

Substituting the known values:
\[ 0 = 30 + (-10) \cdot t \]

### Step 3: Solve for time (t)
Rearranging the equation gives:
\[ 10t = 30 \]
\[ t = \frac{30}{10} = 3 \text{ seconds} \]

This time (3 seconds) is the time taken for the ball to reach the highest point (ascent).

### Step 4: Determine the total time for the ball to return
The time taken to ascend is equal to the time taken to descend. Therefore, the total time (T) for the ball to return to the player's hands is:
\[ T = \text{Time of ascent} + \text{Time of descent} \]
\[ T = 3 \text{ seconds} + 3 \text{ seconds} = 6 \text{ seconds} \]

### Final Answer
The total time taken for the ball to return to the player's hands is **6 seconds**.

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