Question
A ball of mass 5 kg is dropped from a height of 10 m. Find the kinetic energy of the ball just before it hits the ground. (Take g = 10 m/s², no air resistance.)
Solution — Step by Step
Mass m = 5 kg, height h = 10 m, g = 10 m/s². The ball starts from rest, so initial KE = 0. We need KE at the bottom.
Since there’s no air resistance, the total mechanical energy is conserved throughout the fall. All the potential energy at the top converts completely into kinetic energy at the bottom.
At the bottom, height = 0, so PE = 0. At the top, the ball is at rest, so KE = 0.
The potential energy at height h is:
This entire 500 J must go somewhere — and with no friction or air resistance, it all becomes kinetic energy.
By conservation of energy:
Why This Works
Energy conservation is the key idea here. When we say “no air resistance,” we mean no energy is lost to heat or sound — the system is conservative. The ball at 10 m height has a “stored” energy of 500 J simply because of its position. Gravity does work on the ball as it falls, converting that stored energy into motion.
Think of it this way: the height is a kind of “energy account.” As the ball descends, it withdraws from the PE account and deposits into the KE account. At the very bottom, the PE account hits zero and the KE account holds exactly 500 J.
This is why we don’t even need to calculate the velocity to find KE — energy conservation gives us the answer directly, which is much faster in an exam.
Alternative Method
We can also find KE by first calculating the velocity at the bottom, then using KE = ½mv².
Using kinematics: v² = u² + 2gh
Since the ball starts from rest, u = 0:
Now plug into the KE formula:
Same answer. The energy conservation method is faster — one formula instead of two. In CBSE board exams, both approaches fetch full marks, but energy conservation saves you 30 seconds.
In any “find KE just before hitting ground” problem with no air resistance, the shortcut is always KE = mgh. The “just before hitting” phrasing is a signal that you need the full conversion — 100% of PE becomes KE.
Common Mistake
Students often try to find the velocity first using v = √(2gh), calculate it as √200 ≈ 14.14 m/s, then compute ½ × 5 × 200 — and get the right answer, but waste time. The bigger error is using v = √(2gh) and then forgetting to square it properly in ½mv², writing ½ × 5 × 14.14 = 35.35 J. Always use v² directly, not v, in the KE formula.