In the above problem if body is thrown down with velocity 'u' the equation for the descent time is

Options

A. `h=1/2 g t^2`
B. `h=ut+1/2 g t^2`
C. `-h=-ut+1/2 g t^2`
D. `-h=-ut+1/2 g t^2`

Answer

To solve the problem of finding the descent time when a body is thrown down with an initial velocity 'u', we can use the equations of motion under uniform acceleration due to gravity. Here’s the step-by-step solution:

### Step 1: Understand the problem
When a body is thrown downward with an initial velocity 'u', it is subject to gravitational acceleration 'g' (approximately 9.81 m/s²). We need to find the time 't' it takes for the body to descend a certain distance 's'.

### Step 2: Write the equation of motion
The equation of motion that relates initial velocity, acceleration, distance, and time is given by:
\[ s = ut + \frac{1}{2}gt^2 \]
Where:
- \( s \) = distance traveled
- \( u \) = initial velocity (downward)
- \( g \) = acceleration due to gravity (approximately 9.81 m/s²)
- \( t \) = time of descent

### Step 3: Rearrange the equation
To find the time 't', we need to rearrange the equation. This is a quadratic equation in the form of:
\[ \frac{1}{2}gt^2 + ut - s = 0 \]

### Step 4: Apply the quadratic formula
Using the quadratic formula \( t = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \), we identify:
- \( a = \frac{1}{2}g \)
- \( b = u \)
- \( c = -s \)

Substituting these values into the quadratic formula gives:
\[ t = \frac{-u \pm \sqrt{u^2 - 4 \cdot \frac{1}{2}g \cdot (-s)}}{2 \cdot \frac{1}{2}g} \]

### Step 5: Simplify the expression
This simplifies to:
\[ t = \frac{-u \pm \sqrt{u^2 + 2gs}}{g} \]

### Step 6: Choose the correct root
Since time cannot be negative, we take the positive root:
\[ t = \frac{-u + \sqrt{u^2 + 2gs}}{g} \]

### Final Answer
Thus, the equation for the descent time when a body is thrown down with an initial velocity 'u' is:
\[ t = \frac{-u + \sqrt{u^2 + 2gs}}{g} \]

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