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 will follow these steps:

### Step 1: Understand the Motion
The ball is thrown upwards and will eventually come back down to the ground. We need to analyze the motion from the moment it is thrown until it strikes the ground.

### Step 2: Set Up the Equation of Motion
We will use the equation of motion:
\[ s = ut + \frac{1}{2} a t^2 \]
where:
- \( s \) is the displacement (which will be \(-h\) since the ball moves downwards),
- \( u \) is the initial velocity (\( v \), upwards),
- \( a \) is the acceleration (which is \(-g\), downwards),
- \( t \) is the time taken.

### Step 3: Substitute Values into the Equation
Substituting the known values into the equation:
\[ -h = vt - \frac{1}{2} g t^2 \]
Rearranging gives:
\[ \frac{1}{2} g t^2 - vt - h = 0 \]
Multiplying through by 2 to eliminate the fraction:
\[ g t^2 - 2vt - 2h = 0 \]

### Step 4: Identify Coefficients for the Quadratic Formula
This is a quadratic equation in the form \( at^2 + bt + c = 0 \), where:
- \( a = g \),
- \( b = -2v \),
- \( c = -2h \).

### Step 5: Apply the Quadratic Formula
Using the quadratic formula:
\[ t = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \]
Substituting the coefficients:
\[ t = \frac{-(-2v) \pm \sqrt{(-2v)^2 - 4(g)(-2h)}}{2g} \]
This simplifies to:
\[ t = \frac{2v \pm \sqrt{4v^2 + 8gh}}{2g} \]
Further simplifying gives:
\[ t = \frac{2v \pm 2\sqrt{v^2 + 2gh}}{2g} \]
\[ t = \frac{v \pm \sqrt{v^2 + 2gh}}{g} \]

### Step 6: Determine the Valid Solution
Since time cannot be negative, we only consider the positive solution:
\[ t = \frac{v + \sqrt{v^2 + 2gh}}{g} \]

### Final Answer
Thus, the time taken for the ball to strike the ground is:
\[ t = \frac{v}{g} \left(1 + \sqrt{1 + \frac{2gh}{v^2}}\right) \]