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 finding the time taken by a ball thrown upwards with an initial speed \( u \) from a height \( h \) above the ground until it hits the ground, we can break the problem into two parts: the time taken to reach the highest point and the time taken to fall from that highest point to the ground.

### Step-by-step Solution:

1. **Determine the time taken to reach the highest point (T1)**:
- When the ball is thrown upwards, it will eventually stop rising and reach its highest point where its velocity becomes zero.
- Using the equation of motion:
\[ v = u - gT_1 \]
where \( v = 0 \) (the velocity at the highest point), \( u \) is the initial velocity, and \( g \) is the acceleration due to gravity (which is negative as it acts downwards).
- Rearranging gives:
\[ 0 = u - gT_1 \implies gT_1 = u \implies T_1 = \frac{u}{g} \]

2. **Determine the height reached above the initial height (H)**:
- We can use the second equation of motion to find the maximum height reached:
\[ H = uT_1 - \frac{1}{2}gT_1^2 \]
- Substituting \( T_1 = \frac{u}{g} \):
\[ H = u \left(\frac{u}{g}\right) - \frac{1}{2}g\left(\frac{u}{g}\right)^2 \]
\[ H = \frac{u^2}{g} - \frac{1}{2}\frac{u^2}{g} = \frac{u^2}{2g} \]

3. **Total height from the ground**:
- The total height from the ground when the ball reaches its highest point is:
\[ H_{total} = h + H = h + \frac{u^2}{2g} \]

4. **Determine the time taken to fall from the highest point to the ground (T2)**:
- Now, we need to calculate the time taken to fall from height \( H_{total} \) to the ground. Using the second equation of motion:
\[ S = ut + \frac{1}{2}gt^2 \]
- Here, the initial velocity \( u = 0 \) (the ball starts falling from rest at the highest point):
\[ H_{total} = \frac{1}{2}gT_2^2 \]
- Rearranging gives:
\[ T_2^2 = \frac{2H_{total}}{g} = \frac{2(h + \frac{u^2}{2g})}{g} = \frac{2h}{g} + \frac{u^2}{g^2} \]
- Therefore,
\[ T_2 = \sqrt{\frac{2h}{g} + \frac{u^2}{g^2}} \]

5. **Total time taken (T)**:
- The total time taken by the ball to hit the ground is the sum of \( T_1 \) and \( T_2 \):
\[ T = T_1 + T_2 = \frac{u}{g} + \sqrt{\frac{2h}{g} + \frac{u^2}{g^2}} \]

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