A ball is thrown upward with such a velocity v that it returns to the thrower after 3 s. Take g = 10 `ms^(-2)`. Find the value of v :-

Options

A. 15 m/s
B. 20 m/s
C. 10 m/s
D. 5 m/s

Answer

To solve the problem, we need to find the initial velocity \( v \) of a ball thrown upward such that it returns to the thrower after 3 seconds. We will use the equations of motion under uniform acceleration due to gravity.

### Step-by-Step Solution:

1. **Understanding the total time of flight**:
The total time of flight for the ball is given as 3 seconds. When a ball is thrown upwards, it takes equal time to reach its maximum height and to return back to the thrower. Therefore, the time taken to reach the maximum height \( t_1 \) is half of the total time:
\[ t_1 = \frac{3}{2} = 1.5 \text{ seconds} \]

2. **Using the first equation of motion**:
The first equation of motion is given by:
\[ v = u + at \]
where:
- \( v \) is the final velocity,
- \( u \) is the initial velocity,
- \( a \) is the acceleration (which will be negative due to gravity),
- \( t \) is the time.

At the maximum height, the final velocity \( v \) becomes 0 m/s. The acceleration \( a \) due to gravity is \( -g \) (taking upward direction as positive). Therefore, we can substitute the values into the equation:
\[ 0 = v - g t_1 \]

3. **Substituting known values**:
We know \( g = 10 \, \text{m/s}^2 \) and \( t_1 = 1.5 \, \text{s} \):
\[ 0 = v - 10 \times 1.5 \]
Simplifying this gives:
\[ v = 10 \times 1.5 \]
\[ v = 15 \, \text{m/s} \]

4. **Conclusion**:
The initial velocity \( v \) with which the ball was thrown upward is \( 15 \, \text{m/s} \).

### Final Answer:
The value of \( v \) is \( 15 \, \text{m/s} \).