Answer:
The speed of the ball 1.0 m above the ground is 44 m/s (Answer A).
Explanation:
Hi there!
To solve this problem, let´s use the law of conservation of energy. Since there is no air resistance, the only energies that we should consider is the gravitational potential energy and the kinetic energy. Because of the conservation of energy, the loss of potential energy of the ball must be compensated by a gain in kinetic energy.
In this case, the potential energy is being converted into kinetic energy as the ball falls (this is only true when there are no dissipative forces, like air resistance, acting on the ball). Then, the loss of potential energy (PE) is equal to the increase in kinetic energy (KE):
We can express this mathematically as follows:
-ΔPE = ΔKE
-(final PE - initial PE) = final KE - initial KE
The equation of potential energy is the following:
PE = m · g · h
Where:
PE = potential energy.
m = mass of the ball.
g = acceleration due to gravity.
h = height.
The equation of kinetic energy is the following:
KE = 1/2 · m · v²
Where:
KE = kinetic energy.
m = mass of the ball.
v = velocity.
Then:
-(final PE - initial PE) = final KE - initial KE
-(m · g · hf - m · g · hi) = 1/2 · m · v² - 0 (initial KE = 0 because the ball starts from rest) (hf = final height, hi = initial height)
- m · g (hf - hi) = 1/2 · m · v²
2g (hi - hf) = v²
√(2g (hi - hf)) = v
Replacing with the given data:
√(2 · 9.8 m/s²(101 m - 1.0 m)) = v
v = 44 m/s
The speed of the ball 1.0 m above the ground is 44 m/s.
, and Vf and Vi are the final and initial voltages.