A stone is dropped from a height of 45 m on a horizontal level ground. There is horizontal wind blowing due to which horizontal acceleration of the stone becomes `10m//s^(2)`. (Take `g=10m//s^(2))`. The time taken (t) by stone to reach the ground and the net horizontal displacement (x) of the stone form the time it is dropped an till it reaches the ground are respectively

Options

A. `t=3s, x=45m`
B. `t=4s, x=54m`
C. `t=3s, x=54m`
D. `t=4s, x=45m`

Answer

To solve the problem, we need to find the time taken by the stone to reach the ground and the net horizontal displacement of the stone while it falls.

### Step-by-Step Solution:

1. **Identify the Given Data:**
- Height (h) = 45 m
- Acceleration due to gravity (g) = 10 m/s²
- Horizontal acceleration (a_x) = 10 m/s²

2. **Calculate the Time Taken (t) to Reach the Ground:**
The stone is dropped from a height of 45 m. We can use the second equation of motion for vertical displacement:
\[ s = ut + \frac{1}{2} g t^2 \]
Here, the initial velocity (u) is 0 (since the stone is dropped), and the displacement (s) is -45 m (downward). Thus, we have:
\[ -45 = 0 \cdot t + \frac{1}{2} \cdot 10 \cdot t^2 \]
Simplifying this gives:
\[ -45 = 5t^2 \]
\[ t^2 = \frac{-45}{-5} = 9 \]
\[ t = \sqrt{9} = 3 \text{ seconds} \]

3. **Calculate the Horizontal Displacement (x):**
The horizontal displacement can be calculated using the formula:
\[ x = ut + \frac{1}{2} a_x t^2 \]
Since the initial horizontal velocity (u) is 0, we have:
\[ x = 0 \cdot t + \frac{1}{2} \cdot 10 \cdot (3)^2 \]
\[ x = 0 + \frac{1}{2} \cdot 10 \cdot 9 \]
\[ x = 5 \cdot 9 = 45 \text{ meters} \]

### Final Answers:
- Time taken (t) = 3 seconds
- Horizontal displacement (x) = 45 meters