In first second of an object dropped from some height, the distance by which it will fall is `("take g = 10 "ms"^(–2))`

Options

A. 5m
B. 10m
C. 2.5m
D. 15m

Answer

To solve the problem of finding the distance an object falls in the first second after being dropped from a height, we can use the second equation of motion. Here are the steps to arrive at the solution:

### Step 1: Identify the parameters
- Initial velocity (u) = 0 m/s (since the object is dropped)
- Acceleration (a) = g = 10 m/s² (acceleration due to gravity)
- Time (t) = 1 second

### Step 2: Write down the second equation of motion
The second equation of motion is given by:
\[ s = ut + \frac{1}{2} a t^2 \]
where:
- \( s \) = distance traveled
- \( u \) = initial velocity
- \( a \) = acceleration
- \( t \) = time

### Step 3: Substitute the known values into the equation
Since the initial velocity \( u = 0 \):
\[ s = 0 \cdot t + \frac{1}{2} g t^2 \]
Substituting \( g = 10 \, \text{m/s}^2 \) and \( t = 1 \, \text{s} \):
\[ s = 0 + \frac{1}{2} \cdot 10 \cdot (1)^2 \]

### Step 4: Calculate the distance
Now, calculate the value:
\[ s = \frac{1}{2} \cdot 10 \cdot 1 = 5 \, \text{m} \]

### Final Answer
The distance fallen by the object in the first second is **5 meters**.

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