Lifting an object stores gravitational potential energy. Enter the mass and height to get the energy stored and the speed it would reach if released, using conservation of energy.
Enter the mass and height.
Keep g = 9.8 m/s² for Earth.
Read the energy and impact speed.
Potential energy: PE = m g h
Fall speed: v = √(2 g h)
Lift an object and you store energy in it. This calculator finds that gravitational potential energy from the mass, the height and the value of g, using PE = mgh. It then uses conservation of energy to show the speed the object would reach if it were dropped from rest and fell freely through that height.
The idea appears in Class 9 work and energy and again in Class 11. It explains why water stored behind a dam can run turbines, why a falling coconut can hurt, and how much work a lift motor must do to raise a load. Because you can change g, the same tool also shows how the stored energy would differ on the Moon or another planet.
1. Write the mass in kilograms.
2. Measure the height h in metres above your chosen reference level, such as the floor or the ground.
3. Choose g. Use 9.8 m/s² near the Earth's surface unless the problem gives another value.
4. Multiply: PE = m × g × h. The result is in joules.
5. For the fall speed, set the energy lost equal to kinetic energy gained, mgh = ½mv², so v = √(2gh).
6. Remember that v does not depend on mass and that this ignores air resistance.
To lift a mass m slowly through a height h you must push up with a force equal to its weight, mg. The work you do is force times distance, mgh. That work is not lost; it is stored in the Earth-object system as potential energy and can be recovered when the object comes down. This is why the formula has exactly the same form as work, and why potential energy is measured in joules just like kinetic energy.
Height has no natural zero. You can measure it from the floor, the table top or sea level, and the value of PE will change. What stays the same is the difference in potential energy between two positions, and only differences matter physically. If a point is below your reference level, h is negative and so is PE. The calculator accepts negative heights for this reason, but it shows an impact speed of zero because nothing can fall upward from rest.
The formula assumes g is constant over the height involved. That is excellent for buildings, hills and dams, but not for satellites. For large distances from the Earth, the exact expression is U = −GMm ÷ r, which tends to zero far away and becomes more negative closer in. For heights small compared with the Earth's radius of about 6,371 km, the change in −GMm ÷ r reduces to mgh. The impact speed also ignores air resistance, which matters for light or fast objects.
A construction worker in Hyderabad hoists a 40 kg bag of cement to a slab 12 m above the ground.
PE = m g h: = 40 × 9.8 × 12 = 4,704 J
Speed if dropped from rest (no air resistance): v = √(2gh) = 15.3362 m/s
Answer: Potential energy 4,704 J; Impact speed if dropped 15.3362 m/s
Using mass in grams or height in centimetres without converting to kg and m.
Measuring height along a slope instead of the vertical rise.
Believing heavier objects hit the ground faster, when √(2gh) has no mass in it.
Using mgh for a satellite or rocket far above the Earth, where g is much smaller.
Forgetting that real falls lose energy to air resistance, so the true impact speed is lower.
Solving Class 9 and Class 11 work-energy numericals.
Estimating energy stored by water in hydroelectric reservoirs and overhead tanks.
Working out the work a lift, crane or hoist does in raising a load.
Explaining the energy changes in roller coasters, swings and pendulums.
Checking safety of dropped objects on construction sites.
Where is h measured from?
From any reference level you choose; only changes in PE matter.
Does a heavier object fall faster?
Without air resistance, no; √(2gh) does not depend on mass.