A body of mass 1000kg is released from a height of 10m above the ground. Determine its kinetic energy just before it strikes the ground .[g = 10$ms^{-2}$]

Analyze the question: A body falls freely from rest through 10 m. We need its kinetic energy just before it hits the ground.

Concept: With no air resistance, energy is conserved. The potential energy lost becomes kinetic energy:

KE (at ground) = PE (at the top) = mgh

Solution:

• m = 1000 kg
• g = 10 m/s²
• h = 10 m

KE = mgh = 1000 × 10 × 10 = 100,000 J = 10⁵ J

Check using kinematics: v² = u² + 2gh = 0 + 2(10)(10) = 200 KE = ½mv² = ½ × 1000 × 200 = 100,000 J ✔

Answer: The kinetic energy is 10⁵ J. In the options as printed, "10J", "103J" and "104J" are almost certainly 10 J, 10³ J and 10⁴ J, with the powers lost in formatting.

Tip: For any object dropped from rest, KE just before impact = mgh. Quickly multiply the numbers and count the zeros: 1000 × 10 × 10 has five zeros, giving 10⁵. Also watch the formatting of options in WASSCE questions, since powers of 10 are easy to misread.