A ball is projected upwards from a height h above the surface of the earth with velocity v. The time at which the ball strikes the ground is

Options

A. `(v)/(g)+(2hg)/(sqrt2)`
B. `(v)/(g)[1-sqrt(1+(2h)/(g))]`
C. `(v)/(g)[1+sqrt(1+(2gh)/(v^2))]`
D. `(v)/(g)[1+sqrt(v^2+(2h)/(h))]`

Explanation

To solve the problem of determining the time at which a ball projected upwards from a height \( h \) above the surface of the Earth with an initial velocity \( v \) strikes the ground, we can break the motion into two parts: the upward motion until it reaches the highest point and the downward motion from that point to the ground.

### Step-by-Step Solution:

1. **Identify the Two Parts of Motion**:
- **Part 1**: The ball moves upwards until it reaches its highest point.
- **Part 2**: The ball falls from the highest point to the ground.

2. **Calculate Time for Upward Motion (\( t_1 \))**:
- At the highest point, the final velocity \( V = 0 \).
- Initial velocity \( u = v \) (the velocity with which it was projected).
- The acceleration \( a = -g \) (acceleration due to gravity acting downwards).
- Using the first equation of motion:
\[ V = u + at \implies 0 = v - gt_1 \implies t_1 = \frac{v}{g} \]

3. **Calculate the Maximum Height Reached (\( x \))**:
- Using the third equation of motion:
\[ V^2 = u^2 + 2as \implies 0 = v^2 - 2g x \implies x = \frac{v^2}{2g} \]

4. **Calculate Time for Downward Motion (\( t_2 \))**:
- The ball falls from a height \( x + h \) (the maximum height plus the initial height).
- Initial velocity for this part \( u = 0 \) (the ball starts falling from rest).
- Using the second equation of motion:
\[ s = ut + \frac{1}{2} a t^2 \implies -(x + h) = 0 + \frac{1}{2}(-g)t_2^2 \]
\[ -(x + h) = -\frac{1}{2} g t_2^2 \implies t_2^2 = \frac{2(x + h)}{g} \implies t_2 = \sqrt{\frac{2(x + h)}{g}} \]

5. **Substituting for \( x \)**:
- Substitute \( x = \frac{v^2}{2g} \) into the equation for \( t_2 \):
\[ t_2 = \sqrt{\frac{2\left(\frac{v^2}{2g} + h\right)}{g}} = \sqrt{\frac{v^2 + 2gh}{g}} \]

6. **Total Time of Flight (\( T \))**:
- The total time \( T \) is the sum of \( t_1 \) and \( t_2 \):
\[ T = t_1 + t_2 = \frac{v}{g} + \sqrt{\frac{v^2 + 2gh}{g}} \]

7. **Final Expression**:
- Thus, the total time at which the ball strikes the ground is:
\[ T = \frac{v}{g} + \sqrt{\frac{v^2 + 2gh}{g}} \]