A body falls from a height 'h'. In absence of air resistance time of descent of body is

Options

A. `sqrt((2h)/(g))`
B. `sqrt((2h)/(g-a))`
C. `sqrt((2h)/(gpma))`
D. `(h)/(g)`

Answer

To find the time of descent of a body falling from a height \( h \) in the absence of air resistance, we can use the equations of motion. Here's a step-by-step solution:

### Step 1: Identify the known values
- Initial velocity \( u = 0 \) (the body is initially at rest)
- Acceleration due to gravity \( g \) (approximately \( 9.81 \, \text{m/s}^2 \))
- Distance fallen \( s = h \)

### Step 2: Use the equation of motion
We can use the second equation of motion:
\[ s = ut + \frac{1}{2} a t^2 \]
In this case, since the body is falling, we can substitute:
- \( s = h \)
- \( u = 0 \)
- \( a = g \)

So the equation becomes:
\[ h = 0 \cdot t + \frac{1}{2} g t^2 \]
This simplifies to:
\[ h = \frac{1}{2} g t^2 \]

### Step 3: Rearrange the equation to solve for \( t \)
To isolate \( t^2 \), we multiply both sides by 2:
\[ 2h = g t^2 \]
Now, divide both sides by \( g \):
\[ t^2 = \frac{2h}{g} \]

### Step 4: Take the square root to find \( t \)
Now, taking the square root of both sides gives us the time of descent \( t_d \):
\[ t_d = \sqrt{\frac{2h}{g}} \]

### Final Result
Thus, the time of descent of the body falling from a height \( h \) in the absence of air resistance is:
\[ t_d = \sqrt{\frac{2h}{g}} \]

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