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)+ (2g)/(h))]`

Answer

To solve the problem of finding 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 follow these steps:

### Step-by-Step Solution:

1. **Identify Variables**:
- Let \( h \) be the height from which the ball is projected.
- Let \( v \) be the initial velocity of the ball.
- Let \( g \) be the acceleration due to gravity (approximately \( 9.81 \, \text{m/s}^2 \)).
- Let \( t \) be the time taken for the ball to hit the ground.

2. **Set Up the Equation of Motion**:
The equation of motion for the vertical displacement \( s \) can be expressed as:
\[ s = ut + \frac{1}{2} a t^2 \]
Here, \( s = -h \) (the ball moves downwards), \( u = v \) (the initial velocity), and \( a = -g \) (acceleration due to gravity acting downwards). Thus, we can write:
\[ -h = vt - \frac{1}{2} g t^2 \]

3. **Rearranging the Equation**:
Rearranging the equation gives:
\[ \frac{1}{2} g t^2 - vt - h = 0 \]
This is a standard quadratic equation in the form \( at^2 + bt + c = 0 \), where:
- \( a = \frac{1}{2} g \)
- \( b = -v \)
- \( c = -h \)

4. **Using the Quadratic Formula**:
The quadratic formula to find \( t \) 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:
\[ t = \frac{v \pm \sqrt{v^2 + 2gh}}{g} \]

5. **Choosing the Positive Root**:
Since time cannot be negative, we take the positive root:
\[ t = \frac{v + \sqrt{v^2 + 2gh}}{g} \]

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