How long does it take a brick to reach the ground if dropped from a height of 65 m? What will be its velocity just before it reaches the ground ?

Options

A. 2 s , 10 m/s
B. 3.64 s, 35.67 m/s
C. 12 s , 120 m/s
D. 5 s , 20 m/s

Answer

To solve the problem of how long it takes for a brick to reach the ground when dropped from a height of 65 meters, and to find its velocity just before it reaches the ground, we can use the equations of motion under uniform acceleration due to gravity.

### Step-by-step Solution:

1. **Identify the given values:**
- Height (h) = 65 m
- Initial velocity (u) = 0 m/s (since the brick is dropped)
- Acceleration due to gravity (g) = 9.81 m/s² (approximately 10 m/s² for simplicity)

2. **Use the second equation of motion to find the time (t):**
The second equation of motion states:
\[ s = ut + \frac{1}{2}gt^2 \]
Here, \(s\) is the distance fallen (65 m), \(u\) is the initial velocity (0 m/s), and \(g\) is the acceleration due to gravity (approximately 10 m/s²). Plugging in the values:
\[ 65 = 0 \cdot t + \frac{1}{2} \cdot 10 \cdot t^2 \]
This simplifies to:
\[ 65 = 5t^2 \]
Rearranging gives:
\[ t^2 = \frac{65}{5} = 13 \]
Taking the square root:
\[ t = \sqrt{13} \approx 3.61 \text{ seconds} \]

3. **Use the third equation of motion to find the final velocity (v):**
The third equation of motion states:
\[ v^2 = u^2 + 2gh \]
Substituting the known values:
\[ v^2 = 0 + 2 \cdot 10 \cdot 65 \]
This simplifies to:
\[ v^2 = 1300 \]
Taking the square root:
\[ v = \sqrt{1300} \approx 36.06 \text{ m/s} \]

### Final Answers:
- Time taken to reach the ground: **Approximately 3.61 seconds**
- Velocity just before it reaches the ground: **Approximately 36.06 m/s**