Let's break down the problem step-by-step.
1. Calculate the potential energy (PE) at position B:
The formula for potential energy is given by:
[tex]\[
PE = m \times g \times h
\][/tex]
where:
- [tex]\( m \)[/tex] is the mass of the ball, which is 1.5 kilograms.
- [tex]\( g \)[/tex] is the acceleration due to gravity, which is [tex]\( 9.8 \, m/s^2 \)[/tex].
- [tex]\( h \)[/tex] is the height above the ground, which is 0.5 meters.
Substituting the values into the equation:
[tex]\[
PE = 1.5 \, \text{kg} \times 9.8 \, m/s^2 \times 0.5 \, \text{m} = 7.35 \, \text{joules}
\][/tex]
So, the potential energy at position B is [tex]\( 7.35 \)[/tex] joules.
2. Calculate the kinetic energy (KE) at position A:
At position A, all of the potential energy has been converted into kinetic energy, so:
[tex]\[
KE = PE = 7.35 \, \text{joules}
\][/tex]
3. Calculate the velocity of the ball at position A:
The formula for kinetic energy in terms of velocity is:
[tex]\[
KE = \frac{1}{2} \times m \times v^2
\][/tex]
Rearranging to solve for [tex]\( v \)[/tex]:
[tex]\[
v = \sqrt{\frac{2 \times KE}{m}}
\][/tex]
Substituting the values we have:
[tex]\[
v = \sqrt{\frac{2 \times 7.35 \, \text{joules}}{1.5 \, \text{kg}}} = \sqrt{\frac{14.7}{1.5}} = \sqrt{9.8} \approx 3.13 \, m/s
\][/tex]
So, the answers are:
- The ball has [tex]\( 7.35 \)[/tex] joules of potential energy at position B.
- The velocity of the ball at position A is [tex]\( 3.13 \)[/tex] meters per second.
