If a stone is thrown up with a velocity of `9.8 ms^(-1)`, then how much time will it take to come back ?

Options

A. 1 s
B. 2 s
C. 3 s
D. 4 s

Explanation

To solve the problem of how much time a stone will take to come back after being thrown upwards with a velocity of \(9.8 \, \text{m/s}\), we can follow these steps:

### Step 1: Understand the motion
When the stone is thrown upwards, it will rise to a certain height before coming back down. The total displacement of the stone when it returns to the starting point (ground level) is zero.

### Step 2: Identify the initial velocity and acceleration
- Initial velocity (\(u\)) = \(9.8 \, \text{m/s}\) (upward)
- Acceleration due to gravity (\(g\)) = \(9.8 \, \text{m/s}^2\) (downward, hence we will take it as negative when applying the equations of motion)

### Step 3: Use the equation of motion
We can use the equation of motion:
\[ s = ut + \frac{1}{2} a t^2 \]
In this case, since the stone returns to the ground, the displacement (\(s\)) is \(0\). The acceleration \(a\) is \(-g\) (since it acts downward).

Substituting the values, we have:
\[ 0 = ut + \frac{1}{2} (-g) t^2 \]
This simplifies to:
\[ 0 = ut - \frac{1}{2} gt^2 \]

### Step 4: Rearranging the equation
Rearranging gives:
\[ ut = \frac{1}{2} gt^2 \]
Dividing both sides by \(t\) (assuming \(t \neq 0\)):
\[ u = \frac{1}{2} gt \]

### Step 5: Solve for time \(t\)
Now, we can solve for \(t\):
\[ t = \frac{2u}{g} \]
Substituting the values of \(u\) and \(g\):
\[ t = \frac{2 \times 9.8}{9.8} = 2 \, \text{s} \]

### Conclusion
The time taken by the stone to come back to the ground is \(2 \, \text{seconds}\).

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