Answer :
Answer:
To find the speed of the stone after it has fallen a certain distance, we can use the equation for the speed of an object in free fall:
\[ \text{Speed} = \sqrt{2gh} \]
where:
- \( g = 9.8 m/s^2 \) (acceleration due to gravity)
- \( h \) is the distance fallen
Given that \( h = 1m \):
\[ \text{Speed} = \sqrt{2 * 9.8 * 1} = \sqrt{19.6} \approx 4.43 m/s \]
Rounded to two decimal places, the speed of the stone after falling 1m is approximately 4.43 m/s.
Given that \( h = 5m \):
\[ \text{Speed} = \sqrt{2 * 9.8 * 5} = \sqrt{98} \approx 9.90 m/s \]
Rounded to two decimal places, the speed of the stone after falling 5m is approximately 9.90 m/s.
Step-by-step explanation:
To find the speed of the stone when it has fallen a certain distance, we can use the equation for the speed of an object in free fall, which is \( v = \sqrt{2gh} \), where:
- \( v \) is the final speed of the stone (which is given as \( \sqrt{19.6s} \)),
- \( g \) is the acceleration due to gravity (approximately \( 9.8 \, \text{m/s}^2 \)), and
- \( h \) is the height fallen.
Given that \( v = \sqrt{19.6s} \), we can substitute this into the equation to solve for \( s \):
\[ \sqrt{19.6s} = \sqrt{2 \cdot 9.8 \cdot h} \]
Squaring both sides to eliminate the square root gives:
\[ 19.6s = 2 \cdot 9.8 \cdot h \]
\[ 19.6s = 19.6h \]
Now, solve for \( s \) using the given heights \( s = 1 \, \text{m} \) and \( s = 5 \, \text{m} \):
a) For \( s = 1 \, \text{m} \):
\[ 19.6 \cdot 1 = 19.6h \]
\[ h = 1 \, \text{m} \]
b) For \( s = 5 \, \text{m} \):
\[ 19.6 \cdot 5 = 19.6h \]
\[ h = 5 \, \text{m} \]
So, the speed of the stone, correct to 2 decimal places, after it has fallen a distance of \( 1 \, \text{m} \) is \( 1 \, \text{m/s} \) and after it has fallen a distance of \( 5 \, \text{m} \) is \( 5 \, \text{m/s} \).