A 1.0 kg ball falls from height 2.0 m. What is its kinetic energy just before impact?

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Multiple Choice

A 1.0 kg ball falls from height 2.0 m. What is its kinetic energy just before impact?

When something falls, gravity transfers its stored potential energy into kinetic energy (assuming no air resistance). Because energy is conserved, the kinetic energy just before impact equals the initial gravitational potential energy: KE = m g h. Here, m = 1.0 kg, g ≈ 9.8 m/s^2, and h = 2.0 m, so KE = 1 × 9.8 × 2 = 19.6 J. The other numbers would come from different fall heights: 9.8 J if it fell 1 m, 39.2 J if it fell 4 m, and 0 J only if there was no fall. If you want the speed just before impact, v = sqrt(2 g h) ≈ sqrt(39.2) ≈ 6.3 m/s.

When something falls, gravity transfers its stored potential energy into kinetic energy (assuming no air resistance). Because energy is conserved, the kinetic energy just before impact equals the initial gravitational potential energy: KE = m g h.

Here, m = 1.0 kg, g ≈ 9.8 m/s^2, and h = 2.0 m, so KE = 1 × 9.8 × 2 = 19.6 J.

The other numbers would come from different fall heights: 9.8 J if it fell 1 m, 39.2 J if it fell 4 m, and 0 J only if there was no fall. If you want the speed just before impact, v = sqrt(2 g h) ≈ sqrt(39.2) ≈ 6.3 m/s.

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