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 find the time at which the ball strikes the ground after being projected upwards from a height \( h \) with an initial velocity \( v \), we can use the equations of motion. Here’s a step-by-step solution:

### Step 1: Identify the known quantities
- Initial height \( h \)
- Initial velocity \( v \)
- Acceleration due to gravity \( g \) (acting downward)

### Step 2: Set up the equation of motion
We will use the second equation of motion:
\[ S = Ut + \frac{1}{2} a t^2 \]
where:
- \( S \) is the displacement (which will be \( -h \) when the ball hits the ground),
- \( U \) is the initial velocity \( v \),
- \( a \) is the acceleration \( -g \) (since it is acting downward),
- \( t \) is the time of flight.

### Step 3: Substitute the known values into the equation
The equation becomes:
\[ -h = vt - \frac{1}{2} g t^2 \]
Rearranging gives:
\[ \frac{1}{2} g t^2 - vt - h = 0 \]

### Step 4: Rearrange into standard quadratic form
This is a standard quadratic equation of the form \( at^2 + bt + c = 0 \):
\[ \frac{1}{2} g t^2 - vt - h = 0 \]
where:
- \( a = \frac{1}{2} g \),
- \( b = -v \),
- \( c = -h \).

### Step 5: Apply the quadratic formula
The quadratic formula is given by:
\[ t = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \]
Substituting the values of \( a \), \( b \), and \( c \):
\[ t = \frac{-(-v) \pm \sqrt{(-v)^2 - 4 \cdot \frac{1}{2} g \cdot (-h)}}{2 \cdot \frac{1}{2} g} \]
This simplifies to:
\[ t = \frac{v \pm \sqrt{v^2 + 2gh}}{g} \]

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

### Final Answer
Thus, the time at which the ball strikes the ground is:
\[ t = \frac{v + \sqrt{v^2 + 2gh}}{g} \]