A ball is thrown vertically upwards from the top of tower of height `h` with velocity `v` . The ball strikes the ground after time.

Options

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

Explanation

To solve the problem of a ball thrown vertically upwards from the top of a tower of height \( h \) with an initial velocity \( v \), we need to find the total time taken for the ball to strike the ground. We will break down the solution into several steps.

### Step 1: Understanding the Motion of the Ball
When the ball is thrown upwards, it first moves against gravity until it reaches its maximum height. After reaching the maximum height, it will fall back down to the ground. We will denote:
- \( t_1 \): time taken to reach the maximum height
- \( t_2 \): time taken to fall from the maximum height to the ground

### Step 2: Finding \( t_1 \)
At the maximum height, the final velocity \( v_f = 0 \). We can use the first equation of motion:
\[ v_f = u + at \]
Where:
- \( u = v \) (initial velocity)
- \( a = -g \) (acceleration due to gravity, acting downwards)

Setting \( v_f = 0 \):
\[ 0 = v - gt_1 \]
Rearranging gives:
\[ gt_1 = v \quad \Rightarrow \quad t_1 = \frac{v}{g} \]

### Step 3: Finding \( t_2 \)
Now, we need to find the time \( t_2 \) taken to fall from the maximum height back to the ground. The distance fallen is the height of the tower \( h \) plus the height reached above the tower. The height reached can be calculated using the formula:
\[ h_{max} = \frac{v^2}{2g} \]
Thus, the total distance \( s \) from the maximum height to the ground is:
\[ s = h + h_{max} = h + \frac{v^2}{2g} \]

Using the second equation of motion for falling:
\[ s = ut + \frac{1}{2} a t^2 \]
Where:
- \( u = 0 \) (initial velocity at the maximum height)
- \( a = g \)

Substituting in gives:
\[ h + \frac{v^2}{2g} = 0 + \frac{1}{2} g t_2^2 \]
Rearranging this gives:
\[ g t_2^2 = 2 \left( h + \frac{v^2}{2g} \right) \]
\[ t_2^2 = \frac{2h}{g} + \frac{v^2}{g^2} \]
Taking the square root:
\[ t_2 = \sqrt{\frac{2h}{g} + \frac{v^2}{g^2}} \]

### Step 4: Total Time of Flight
The total time \( T \) taken for the ball to strike the ground is the sum of the time taken to reach maximum height and the time taken to fall to the ground:
\[ T = t_1 + t_2 = \frac{v}{g} + \sqrt{\frac{2h}{g} + \frac{v^2}{g^2}} \]

### Step 5: Final Expression
To simplify, we can factor out \( \frac{1}{g} \):
\[ T = \frac{1}{g} \left( v + \sqrt{2hg + v^2} \right) \]

### Conclusion
The total time taken for the ball to strike the ground after being thrown from the top of the tower is:
\[ T = \frac{1}{g} \left( v + \sqrt{2hg + v^2} \right) \]