A ball is thrown upwards from the top of a tower `40 m` high with a velocity of `10 m//s.` Find the time when it strikes the ground. Take `g=10 m//s^2`.

Options

A. `4s`
B. `-28s`
C. `12s`
D. `-2s`

Explanation

To solve the problem of finding the time when a ball thrown upwards from the top of a 40 m high tower strikes the ground, we can follow these steps:

### Step 1: Identify the known values
- Initial height of the tower (h) = 40 m
- Initial velocity (u) = 10 m/s (upwards)
- Acceleration due to gravity (g) = 10 m/s² (downwards)

### Step 2: Define the direction of motion
Since the ball is thrown upwards, we will take the upward direction as positive. Therefore, the acceleration due to gravity will be negative:
- Acceleration (a) = -g = -10 m/s²

### Step 3: Set up the equation of motion
We will use the second equation of motion:
\[ S = ut + \frac{1}{2} a t^2 \]
Where:
- S = displacement (final position - initial position)
- u = initial velocity
- a = acceleration
- t = time

### Step 4: Determine the displacement
The ball is thrown from the top of the tower and strikes the ground. The displacement (S) will be negative since it moves downward from the initial position:
\[ S = -40 \, \text{m} \]

### Step 5: Substitute the known values into the equation
Substituting the values into the equation:
\[ -40 = 10t + \frac{1}{2}(-10)t^2 \]
This simplifies to:
\[ -40 = 10t - 5t^2 \]

### Step 6: Rearrange the equation
Rearranging gives us:
\[ 5t^2 - 10t - 40 = 0 \]
Dividing the entire equation by 5:
\[ t^2 - 2t - 8 = 0 \]

### Step 7: Solve the quadratic equation
We can use the quadratic formula:
\[ t = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \]
Where:
- a = 1, b = -2, c = -8

Calculating the discriminant:
\[ b^2 - 4ac = (-2)^2 - 4(1)(-8) = 4 + 32 = 36 \]

Now substituting into the quadratic formula:
\[ t = \frac{2 \pm \sqrt{36}}{2} \]
\[ t = \frac{2 \pm 6}{2} \]

This gives us two possible values for t:
1. \( t = \frac{8}{2} = 4 \) seconds
2. \( t = \frac{-4}{2} = -2 \) seconds (not physically possible)

### Step 8: Conclusion
The time when the ball strikes the ground is:
\[ t = 4 \, \text{seconds} \]