A carpenter on top of a roof of 20.0m high dropped a hammer of mass 1.5 kg and it fell directly to the ground…

Analyze the question: A hammer is dropped (so it starts from rest) from a height of 20.0 m. We need its kinetic energy just before it hits the ground.

Concept: Energy is conserved when we ignore air resistance. The gravitational potential energy at the top changes entirely into kinetic energy at the bottom:

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

Solution:

Given: m = 1.5 kg, g = 10 m/s², h = 20.0 m

KE = mgh KE = 1.5 × 10 × 20 KE = 15 × 20 KE = 300 J

Check using kinematics:

• v² = u² + 2gh = 0 + 2(10)(20) = 400, so v = 20 m/s
• KE = ½mv² = ½ × 1.5 × 400 = 300 J ✔

Answer: The kinetic energy just before hitting the ground is 300 J, which is option C.

Tip: For any "dropped from a height" question, skip the velocity calculation and go straight to KE = mgh. Watch out for the common mistake of forgetting to multiply by the full height. Option B (150 J) comes from wrongly using ½mgh.