Physics · Gravitation · NEET
Take the ground as the PE reference and drop from height S. Total energy at the top is mgS (all PE, no KE). At the height h where KE = 3 PE, total energy = KE + PE = 3 PE + PE = 4 PE = 4(mgh). Set 4mgh = mgS, so h = S/4. That means the body has already fallen 3S/4. This is exactly the NEET 2021 question. The trick is that '3 times' does not mean 3/4 of the height; write total = KE + PE and let the numbers do the work.
Yes, as long as we ignore air resistance. Gravity is a conservative force, so mechanical energy (KE + PE) is conserved. As the body falls, PE decreases and KE increases by the exact same amount, so their sum is constant. If air drag acts, some energy leaves as heat, and KE + PE slowly drops. For NEET, unless drag is stated, treat the sum as constant.
Use PE = mgh only for small heights near Earth's surface (h much smaller than Earth's radius R), where g is nearly constant. Use the full PE = -GMm/r when the height is large, comparable to R, such as escape-velocity or maximum-height-far-from-Earth problems. The full form takes PE = 0 at infinity, so PE is always negative and grows toward 0 as you go higher.
They use different zero points. In PE = -GMm/r the zero is chosen at infinity, so any point closer than infinity has less energy, meaning negative PE. In PE = mgh the zero is chosen at the ground, so points above it are positive. Both describe the same physics; only the reference level differs. What matters in problems is the change in PE, which comes out the same.
KE is maximum at the lowest point of the fall, just before impact, because that is where PE is smallest (most negative or lowest mgh). For a body thrown up, KE is maximum at the launch point and zero at the top. Simple rule: KE peaks where PE is lowest, and KE is zero where the body momentarily stops.
A particle is released from rest at height S above the earth's surface. At a certain height its kinetic energy is three times its potential energy (taking the surface as reference). The height above the surface and the speed at that instant are respectively:
Try the real previous-year questions from this chapter — each with the answer and a full solution.
Their sum is constant (KE + PE = total mechanical energy). When the body rises, KE turns into PE; when it falls, PE turns into KE. The trade is one-for-one if air resistance is ignored.
For a body thrown vertically, KE at the top is zero because the body momentarily stops. All the launch KE has become PE. If thrown at an angle, KE is not zero at the top because the horizontal speed remains.
Yes, if only gravity acts. Gravity is conservative, so KE + PE stays constant. Air resistance breaks this and slowly reduces the total energy as heat.
Set loss in PE equal to gain in KE. Near the surface: mgh = 1/2 m v^2, so v = sqrt(2gh). This avoids using time and is faster than kinematics for NEET.
Gravitation energy questions appear almost every year, and the KE-PE ratio type (like NEET 2021) is a common trap. Mastering 'total = KE + PE first' saves time and prevents the 3S/4 vs S/4 mistake.
