A ball is thrown upwards with a speed u from a height h above the ground.The time taken by the ball to hit the ground is

Options

A. `sqrt2h//g`
B. `sqrt(8h//g`
C. `sqrt(u^(2)+2gh)/(g)`
D. `(u)/(g)+sqrt(2h)/(g)`

Answer

To solve the problem of a ball thrown upwards from a height \( h \) with an initial speed \( u \), we need to determine the total time \( T \) taken for the ball to hit the ground. We can break this problem into two parts: the time taken to reach the maximum height and the time taken to fall from that height to the ground.

### Step-by-Step Solution:

1. **Determine the time to reach maximum height (T1)**:
- When the ball is thrown upwards, it will eventually stop rising and start falling back down. At the maximum height, the final velocity \( V \) is 0.
- Using the equation of motion:
\[ V = u - g t_1 \]
where \( g \) is the acceleration due to gravity (acting downwards).
- Setting \( V = 0 \):
\[ 0 = u - g t_1 \]
- Rearranging gives:
\[ t_1 = \frac{u}{g} \]

2. **Determine the height reached above the initial height (h)**:
- The height reached by the ball above the initial height \( h \) can be calculated using the equation:
\[ s = ut_1 - \frac{1}{2} g t_1^2 \]
- Substituting \( t_1 = \frac{u}{g} \):
\[ s = u \left(\frac{u}{g}\right) - \frac{1}{2} g \left(\frac{u}{g}\right)^2 \]
- Simplifying gives:
\[ s = \frac{u^2}{g} - \frac{1}{2} \frac{u^2}{g} = \frac{u^2}{2g} \]

3. **Calculate the total height from which the ball falls (H)**:
- The total height from which the ball will fall to the ground is:
\[ H = h + s = h + \frac{u^2}{2g} \]

4. **Determine the time to fall from height H (T2)**:
- Using the equation of motion for free fall:
\[ H = \frac{1}{2} g t_2^2 \]
- Rearranging gives:
\[ t_2^2 = \frac{2H}{g} = \frac{2(h + \frac{u^2}{2g})}{g} \]
- Simplifying gives:
\[ t_2^2 = \frac{2h}{g} + \frac{u^2}{g^2} \]
- Taking the square root:
\[ t_2 = \sqrt{\frac{2h}{g} + \frac{u^2}{g^2}} \]

5. **Calculate the total time (T)**:
- The total time taken for the ball to hit the ground is:
\[ T = t_1 + t_2 = \frac{u}{g} + \sqrt{\frac{2h}{g} + \frac{u^2}{g^2}} \]

### Final Answer:
The total time taken by the ball to hit the ground is:
\[ T = \frac{u}{g} + \sqrt{\frac{2h}{g} + \frac{u^2}{g^2}} \]