Find the time taken for the ball to strike the ground . If a ball is thrown straight upwards with a speed v from a point h meters above the ground .

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]`

Answer

To find the time taken for the ball to strike the ground when thrown straight upwards with an initial speed \( v \) from a height \( h \) above the ground, we can use the second equation of motion:

\[ s = ut + \frac{1}{2} a t^2 \]

### Step 1: Define the Variables
- Let \( u = v \) (initial velocity, upwards)
- Let \( h \) be the height from which the ball is thrown.
- 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 strike the ground.

### Step 2: Set Up the Equation
Since the ball is thrown upwards and then falls downwards, we can take downward direction as positive. The displacement \( s \) when the ball strikes the ground will be \( -h \) (as it moves from height \( h \) to ground level). Thus, we can write:

\[ -h = vt - \frac{1}{2} g t^2 \]

Rearranging gives us:

\[ \frac{1}{2} g t^2 - vt - h = 0 \]

### Step 3: Use the Quadratic Formula
This is a quadratic equation in the form \( at^2 + bt + c = 0 \), where:
- \( a = \frac{1}{2} g \)
- \( b = -v \)
- \( c = -h \)

Using the quadratic formula \( t = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \):

\[ t = \frac{-(-v) \pm \sqrt{(-v)^2 - 4 \cdot \frac{1}{2} g \cdot (-h)}}{2 \cdot \frac{1}{2} g} \]

This simplifies to:

\[ t = \frac{v \pm \sqrt{v^2 + 2gh}}{g} \]

### Step 4: Determine the Valid Solution
Since time cannot be negative, we will only consider the positive root:

\[ t = \frac{v + \sqrt{v^2 + 2gh}}{g} \]

### Final Answer
The time taken for the ball to strike the ground is:

\[ t = \frac{v + \sqrt{v^2 + 2gh}}{g} \]

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