Physics · Motion In A Straight Line · NEET
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.
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.
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.
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.
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).
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.
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 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:
Try the real previous-year questions from this chapter — each with the answer and a full solution.
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.
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).
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.
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.
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.
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.
