To solve this problem, we need to determine the speed of the ball when it reaches the base of the hill. This can be done using principles of energy conservation or kinematic equations. Here, we will use the concept of gravitational potential energy converting into kinetic energy.
1. Gravitational Potential Energy at the Top:
- The gravitational potential energy at the top of the hill is given by the formula [tex]\( PE = mgh \)[/tex],
where:
- [tex]\( m \)[/tex] is the mass of the ball (10 kg),
- [tex]\( g \)[/tex] is the acceleration due to gravity (10 m/s²),
- [tex]\( h \)[/tex] is the height of the hill (10 m).
Plugging in the values, we get:
[tex]\[
PE = 10 \, \text{kg} \times 10 \, \text{m/s}^2 \times 10 \, \text{m} = 1000 \, \text{J}
\][/tex]
2. Kinetic Energy at the Base:
- When the ball reaches the base of the hill, all the gravitational potential energy will have converted into kinetic energy.
- The kinetic energy [tex]\( KE \)[/tex] is given by the formula [tex]\( KE = \frac{1}{2} mv^2 \)[/tex],
where [tex]\( v \)[/tex] is the velocity we need to find.
Setting [tex]\( KE = PE \)[/tex], we have:
[tex]\[
\frac{1}{2} mv^2 = mgh
\][/tex]
3. Solving for Velocity:
- Cancel the mass [tex]\( m \)[/tex] from both sides of the equation:
[tex]\[
\frac{1}{2} v^2 = gh
\][/tex]
- Solving for [tex]\( v \)[/tex], we get:
[tex]\[
v^2 = 2gh
\][/tex]
- Taking the square root of both sides:
[tex]\[
v = \sqrt{2gh}
\][/tex]
4. Substitute the Values:
- Substituting [tex]\( g = 10 \, \text{m/s}^2 \)[/tex] and [tex]\( h = 10 \, \text{m} \)[/tex], we get:
[tex]\[
v = \sqrt{2 \times 10 \, \text{m/s}^2 \times 10 \, \text{m}}
\][/tex]
- Simplify inside the square root:
[tex]\[
v = \sqrt{200 \, \text{m}^2/\text{s}^2}
\][/tex]
- Calculate the square root:
[tex]\[
v \approx 14.142135623730951 \, \text{m/s}
\][/tex]
5. Rounding to the Nearest Tenth:
- Finally, rounding 14.142135623730951 to the nearest tenth gives:
[tex]\[
v \approx 14.1 \, \text{m/s}
\][/tex]
Hence, the ball is traveling at approximately [tex]\( 14.1 \, \text{m/s} \)[/tex] when it reaches the base of the hill.
