Question

A basketball weighing 0.63 kg is dropped from a height of 6.0 meters onto a court. Use the conservation of energy equation to determine the velocity of the ball at a height of 2.0 meters above the court.

Answer 1

Answer:

The velocity of the ball at a height of 2.0 meters above the court is approximately 8.85 m/s

Explanation:

The given parameters of the ball are;

The mass of the ball, m = 0.63 kg

The height from which the ball is dropped, h₁ = 6.0 meters

The height at which the velocity of the ball is sought, h₂ = 2 meters

The initial potential energy of the ball, P.E. = m·g·h₁ = 0.63 × 9.8 × 6.0  = 37.044

The initial potential energy of the ball, P.E.₁ = 37.044 J

The potential energy of the ball, when the ball is at 2 meters above the court, P.E.₂ = m·g·h₂ = 0.63 × 9.8 × 2.0 = 12.348

The potential energy of the ball, when the ball is at 2 meters above the court, P.E.₂ = 12.348 J

From M.E> = P.E. + K.E.

Where;

M.E = The total mechanical energy of the ball = Constant

P.E. = The potential energy of the ball

K.E. = The kinetic energy of the ball

By the conservation of energy principle, we have;

The potential energy lost by the ball = The kinetic energy gained by the ball

The potential energy lost by the ball = P.E.₁ - P.E.₂ = 37.044 - 12.348 = 24.696

The potential energy lost by the ball = 24.696 J

The kinetic energy gained by the ball = 1/2·m·v² = 1/2×0.63×v²

Where;

v = The velocity of the ball

∴ The potential energy lost by the ball at 2.0 meters above the court = 24.696 J = The kinetic energy gained by the ball at 2.0 meters above the court = 1/2×0.63×v²

24.696 J = 1/2×0.63 kg ×v²

v² = 24.696 J / (1/2×0.63 kg) = 78.4 m²/s²

∴ v = √(78.4 m²/s²) = 8.85437744847 m/s

The velocity of the ball at a height of 2.0 meters above the court, v ≈ 8.85 m/s.