A body falls from 80 m. Its time of descent is `[g = 10 ms^(-2)]`

Options

A. 3 s
B. 4 s
C. 5s
D. 6s

Answer

To solve the problem of a body falling from a height of 80 meters with a gravitational acceleration of \( g = 10 \, \text{m/s}^2 \), we can use the equations of motion. Here’s a step-by-step solution:

### Step 1: Identify the known values
- Height (\( s \)) = 80 m
- Initial velocity (\( u \)) = 0 m/s (since the body is falling from rest)
- Acceleration (\( a \)) = \( g = 10 \, \text{m/s}^2 \)

### Step 2: Use the equation of motion
We will use the second equation of motion:
\[ s = ut + \frac{1}{2} a t^2 \]
Substituting the known values into the equation:
\[ 80 = 0 \cdot t + \frac{1}{2} \cdot 10 \cdot t^2 \]
This simplifies to:
\[ 80 = 5t^2 \]

### Step 3: Solve for \( t^2 \)
To isolate \( t^2 \), we rearrange the equation:
\[ t^2 = \frac{80}{5} \]
Calculating the right side gives:
\[ t^2 = 16 \]

### Step 4: Find \( t \)
Now, take the square root of both sides to find \( t \):
\[ t = \sqrt{16} = 4 \, \text{seconds} \]

### Conclusion
The time of descent for the body is \( t = 4 \) seconds.

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