A ball released from rest from the roof of a building of height 45 m at t = 0. Find the height of ball from ground at t = 2s.

Options

A. 10 m
B. 20 m
C. 25 m
D. 30 m

Explanation

To find the height of the ball from the ground at \( t = 2 \) seconds, we can use the equations of motion. Here’s a step-by-step solution:

### Step 1: Identify the given information
- Initial height of the building, \( H = 45 \) m
- Initial velocity of the ball, \( u = 0 \) m/s (since the ball is released from rest)
- Acceleration due to gravity, \( g = 9.8 \) m/s² (we can approximate this to \( 10 \) m/s² for simplicity)
- Time, \( t = 2 \) s

### Step 2: Use the equation of motion to find the distance fallen by the ball
We can use the second equation of motion:
\[ s = ut + \frac{1}{2} a t^2 \]
Where:
- \( s \) is the distance fallen
- \( u \) is the initial velocity (0 m/s)
- \( a \) is the acceleration (which is \( g \))
- \( t \) is the time (2 s)

Substituting the values:
\[ s = 0 \cdot 2 + \frac{1}{2} \cdot 10 \cdot (2^2) \]
\[ s = 0 + \frac{1}{2} \cdot 10 \cdot 4 \]
\[ s = 20 \text{ m} \]

### Step 3: Calculate the height of the ball from the ground
The height of the ball from the ground at \( t = 2 \) seconds can be found by subtracting the distance fallen from the initial height of the building:
\[ \text{Height from ground} = H - s \]
Substituting the values:
\[ \text{Height from ground} = 45 \text{ m} - 20 \text{ m} \]
\[ \text{Height from ground} = 25 \text{ m} \]

### Final Answer:
The height of the ball from the ground at \( t = 2 \) seconds is **25 meters**.

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