A ball is thrown straight upward with a speed v from a point h meter above the ground. The time taken for the ball to strike the ground is

Options

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

Explanation

To find the time taken for a ball thrown straight upward with an initial speed \( v \) from a height \( h \) above the ground to strike the ground, we can break the problem into two parts: the time taken to reach the maximum height and the time taken to fall back to the ground.

### Step-by-Step Solution:

1. **Determine the maximum height reached by the ball:**
- When the ball is thrown upwards with speed \( v \), it will rise until its velocity becomes zero.
- Using the kinematic equation:
\[ v^2 = u^2 + 2as \]
where \( v = 0 \) (final velocity at the maximum height), \( u = v \) (initial velocity), \( a = -g \) (acceleration due to gravity), and \( s = h_1 \) (height gained).
- Rearranging gives:
\[ 0 = v^2 - 2gh_1 \implies h_1 = \frac{v^2}{2g} \]

2. **Calculate the total height from which the ball falls:**
- The total height from which the ball will fall to the ground is the initial height \( h \) plus the height gained \( h_1 \):
\[ H = h + h_1 = h + \frac{v^2}{2g} \]

3. **Determine the time taken to fall from height \( H \):**
- For the downward motion from height \( H \), we can use the kinematic equation:
\[ s = ut + \frac{1}{2}at^2 \]
where \( s = H \), \( u = 0 \) (initial velocity at the peak), \( a = g \), and \( t = t_2 \) (time taken to fall).
- Thus, we have:
\[ H = 0 \cdot t_2 + \frac{1}{2}gt_2^2 \implies H = \frac{1}{2}gt_2^2 \implies t_2^2 = \frac{2H}{g} \]
- Substituting \( H \):
\[ t_2^2 = \frac{2(h + \frac{v^2}{2g})}{g} = \frac{2h}{g} + \frac{v^2}{g^2} \]
- Taking the square root gives:
\[ t_2 = \sqrt{\frac{2h}{g} + \frac{v^2}{g^2}} \]

4. **Determine the time taken to reach the maximum height:**
- Using the equation \( v = u + at \):
\[ 0 = v - gt_1 \implies t_1 = \frac{v}{g} \]

5. **Calculate the total time taken to strike the ground:**
- The total time \( T \) is the sum of the time taken to reach the maximum height and the time taken to fall back to the ground:
\[ T = t_1 + t_2 = \frac{v}{g} + \sqrt{\frac{2h}{g} + \frac{v^2}{g^2}} \]

### Final Answer:
\[ T = \frac{v}{g} + \sqrt{\frac{2h}{g} + \frac{v^2}{g^2}} \]