Object Dropped from a Height: Free Fall Calculations
Physics · Motion In A Straight Line · NEET
When an object is simply dropped (released, not thrown), its initial velocity u = 0 and it falls under gravity with a = g (about 9.8 m/s², often 10 m/s² in NEET). Use h = ½gt² for the height, v = gt for the speed after time t, and v² = 2gh to link speed with height. Memory hook: "Dropped means u = 0" — the only thing acting is gravity, so just plug u = 0 into the three equations of motion.
A dropped object starts from rest (u = 0) and accelerates downward at g. Time and final speed depend only on the height h and g, never on mass: h = ½gt², v = gt, v² = 2gh.
Your doubts, answered
When an object is dropped, is the initial velocity really zero?
Yes. "Dropped" or "released" means the object starts from rest, so u = 0 at the instant you let go. This is the single most important fact for these problems. Only when the object is "thrown" (given a push down or up) is u not zero. Once you write u = 0, the equations shrink: v = gt, h = ½gt², and v² = 2gh.
Does a heavier object fall faster than a lighter one?
No. In free fall (air resistance neglected, as NEET assumes), all objects fall with the same acceleration g, so a heavy stone and a light stone dropped from the same height reach the ground at the same time and with the same speed. Mass never appears in t = √(2h/g) or v = √(2gh). This is why a coin and a feather fall together in a vacuum.
What acceleration value should I use, 9.8 or 10?
Read the question. If it says g = 10 m/s² use 10; if it says g = 9.8 m/s² use 9.8. If nothing is given in a NEET numerical, 9.8 m/s² is the standard value, but 10 makes the arithmetic clean. Always take g as positive and downward here, because for a dropped object motion and acceleration are both downward, so you can treat everything as positive.
How do I find how long the object takes to hit the ground?
Use h = ½gt² and solve for t: t = √(2h/g). Example: a stone dropped from h = 45 m with g = 10 gives t = √(2×45/10) = √9 = 3 s. Note t depends only on height and g, not on mass.
How do I find the speed just before it lands?
Two easy ways. If you know the time: v = gt. If you know the height: v = √(2gh). For h = 45 m, g = 10: v = √(2×10×45) = √900 = 30 m/s. Both methods must agree (v = gt = 10×3 = 30 m/s).
The object is dropped from a moving lift or plane, does u stay zero?
Relative to the ground, no — the object shares the vehicle's velocity at release. But relative to the vehicle (if it moves with uniform velocity), the object still falls with u = 0 and a = g, so the time to reach the floor is the same as when the lift is at rest. This exact idea was tested in NEET 2019.
⚠️ The NEET trap ✗ Students carry over the thrown-up sign convention and write a = -g with a negative height, or wrongly assume the object needs some starting speed and plug in a non-zero u. ✓ For a purely dropped object, motion and acceleration are both downward. Take down as positive: u = 0, a = +g, h positive. Then h = ½gt², v = gt, v² = 2gh with all-positive numbers. Only mix in negative signs when the object is also thrown upward first. 🧠 Sign of g and the "dropped means u = 0" rule
Real NEET questions
NEET 2019
A person in a lift drops a coin. The coin takes time t1 to reach the floor when the lift is at rest, and time t2 when the lift is moving up with uniform velocity. Then:
A · t1 = t2 ✓
B · t1 < t2
C · t1 > t2
D · t1 = 2t2
Solution: Uniform velocity means the lift is a non-accelerating (inertial) frame, so there is no extra pseudo-force. Relative to the lift floor the coin still starts from rest (u = 0 relative to the lift) and falls the same height h with the same acceleration g. From h = ½gt², t = √(2h/g), which is identical in both cases. Therefore t1 = t2. Correct answer: A.
NEET 2026
A ruler is dropped vertically and five persons try to catch it. Their reaction times are A = 0.20 s, B = 0.22 s, C = 0.18 s, D = 0.19 s and E = 0.21 s. The correct order of the distance travelled by the ruler before being caught by each person is:
A · C > D > A > E > B
B · A > B > C > D > E
C · B > E > A > D > C ✓
D · B > A > E > C > D
Solution: The ruler is dropped, so u = 0 and the distance fallen is s = ½gt², which increases with reaction time t. Larger t means larger distance. Order the times from largest to smallest: B(0.22) > E(0.21) > A(0.20) > D(0.19) > C(0.18). The distances follow the same order: B > E > A > D > C. Correct answer: C.
Solved Motion In A Straight Line NEET PYQs
Try the real previous-year questions from this chapter — each with the answer and a full solution.
What is the initial velocity of an object dropped from a height?
Zero. A dropped or released object starts from rest, so u = 0. It then speeds up because of gravity, gaining g (about 9.8 m/s²) of speed every second.
What is the formula for the height from which an object is dropped?
h = ½gt², where g is the acceleration due to gravity and t is the time of fall. Rearranged, the time to fall is t = √(2h/g).
How do you calculate the velocity of a dropped object when it hits the ground?
Use v = gt if you know the fall time, or v = √(2gh) if you know the height. For a 20 m drop with g = 10, v = √(2×10×20) = 20 m/s.
Do two objects of different mass dropped together land at the same time?
Yes, provided air resistance is ignored. Free fall acceleration g is the same for all masses, so both land together with the same speed. Mass cancels out of every free-fall equation.
Is g positive or negative for a dropped object?
Take g as positive when you choose downward as the positive direction, which is the simplest for a purely dropped object. All quantities (u = 0, v, h, a = g) are then positive.
How much distance does a dropped object cover in the first second?
With g = 9.8 m/s², h = ½ × 9.8 × 1² = 4.9 m in the first second. With g = 10, it is 5 m. Successive-second distances follow the 1 : 3 : 5 ratio.
