To determine the velocity at which an object hits the ground when dropped from a height of 43.8 meters, we can utilize the principles of physics, specifically the equations of motion under the influence of gravity.
### Step-by-Step Solution:
1. Identify the given values:
- Height ([tex]\( h \)[/tex]) = 43.8 meters
- Acceleration due to gravity ([tex]\( g \)[/tex]) = 9.81 meters per second squared
- Initial velocity ([tex]\( u \)[/tex]) = 0 meters per second (since the object is dropped)
2. Use the appropriate equation of motion:
The equation we need is:
[tex]\[
v^2 = u^2 + 2gh
\][/tex]
Where [tex]\( v \)[/tex] is the final velocity we need to find.
- Plugging in the known values:
[tex]\[
v^2 = 0^2 + 2 \cdot 9.81 \cdot 43.8
\][/tex]
3. Calculate the product of constants:
- Calculate [tex]\( 2 \cdot 9.81 \cdot 43.8 \)[/tex]:
[tex]\[
v^2 = 2 \cdot 9.81 \cdot 43.8 = 859.356
\][/tex]
4. Find the square root to solve for [tex]\( v \)[/tex]:
- Taking the square root of 859.356:
[tex]\[
v = \sqrt{859.356} \approx 29.314774432016357
\][/tex]
5. Round the final velocity to the nearest tenth:
- Rounding [tex]\( 29.314774432016357 \)[/tex] to the nearest tenth:
[tex]\[
v \approx 29.3 \, \text{meters per second}
\][/tex]
So, the object hits the ground with a velocity of approximately 29.3 meters per second when dropped from a height of 43.8 meters.
