if a stone dropped into a well reaches the water's surface after 3.0 seconds, how far did the stone drop before hitting the water?
S=ut+0.5 gt^2
=0+0.5x9.8x3^2
=44 m
thx
I'm taking the same test as this Jason kid was 3 years ago. I passed ofc😎
To calculate the distance the stone dropped before hitting the water, we need to use the equation of motion for free-falling objects:
```
d = (1/2) * g * t^2
```
Where:
- `d` is the distance traveled
- `g` is the acceleration due to gravity (approximately 9.8 m/s^2 on Earth)
- `t` is the time taken
In this case, we know the time taken is 3.0 seconds. So, substituting the values into the equation, we get:
```
d = (1/2) * 9.8 * (3.0)^2
```
Now, let's calculate that:
```
d = (1/2) * 9.8 * 9.0
d = 44.1 meters
```
Therefore, the stone dropped approximately 44.1 meters before hitting the water surface.