Physics · Motion In A Straight Line · NEET
No. This is the single most common mistake. At the top the VELOCITY is zero for just an instant, but the ACCELERATION is still g = 9.8 m/s² pointing down. If acceleration were zero at the top, the ball would freeze in mid-air forever. Gravity never switches off. Velocity being zero only means the ball has finished going up and is about to start coming down. Acceleration stays constant (g, downward) for the entire flight — going up, at the top, and coming down.
Fix one direction as positive and stick to it for the whole problem. The usual choice: take UP as positive. Then the throw velocity u is positive, but g is NEGATIVE (a = -g = -9.8 m/s²) because gravity points down. So the equations become v = u - gt, H = u²/(2g) as a magnitude, and s = ut - (1/2)g t². If instead you take DOWN as positive, then a = +g and the initial u is negative. Both give the same answer; just never mix the two mid-problem.
At the top the velocity is zero. Use v = u - gt with v = 0: 0 = u - g·t, so t(up) = u/g. Example: u = 20 m/s, g = 10 m/s² gives t = 20/10 = 2 s to reach the top. The time to fall back down is the SAME 2 s, so total time of flight = 2u/g = 4 s.
Yes, if we ignore air resistance. It leaves your hand at speed u and returns to the same height at speed u — same magnitude, opposite direction. This is because the motion is symmetric: the path up is a mirror image of the path down. Use v² = u² - 2g·s: when it returns to the launch point s = 0, so v² = u², meaning |v| = u. Only the direction (sign) is reversed.
Maximum height H = u²/(2g) is the highest point reached, measured from the launch point. But when the ball comes back to your hand, its total DISPLACEMENT is zero (it ended where it started), even though the total DISTANCE travelled is 2H (up H, then down H). NEET often tests this distance-vs-displacement trap in free-fall questions.
On the way up and on the way down, at any given height the ball has the same SPEED but opposite VELOCITY direction. Speed is a scalar (magnitude only), velocity is a vector. At height h going up, velocity is +v; at the same height h coming down, velocity is -v. Same number, opposite sign.
A ball is thrown vertically upward with a velocity of 4 m/s from a bridge. The ball strikes the water surface after 4 s. The height of the bridge above the water surface is (g = 10 m/s²):
A ball is thrown vertically upward and falls back to the thrower. Which velocity (v)–time (t) graph correctly represents its motion? (upward taken positive)
Try the real previous-year questions from this chapter — each with the answer and a full solution.
H = u²/(2g), where u is the launch speed and g ≈ 9.8 m/s² (use 10 m/s² if the question says so). It comes from v² = u² - 2gH with v = 0 at the top.
Total time of flight T = 2u/g. Time going up (t = u/g) equals time coming down, so the total is twice that.
It returns with the same magnitude of speed, u, but directed downward. If up was positive, the return velocity is -u.
No. Velocity is zero at the highest point, but acceleration equals g = 9.8 m/s² downward for the entire motion. Gravity never stops acting.
Total distance = 2H = u²/g (up H, then down H). Total displacement = 0, because it ends at the starting point.
Choose a positive direction and keep it. If up is positive, g is negative (a = -9.8 m/s²). If down is positive, g is positive (a = +9.8 m/s²). Answers come out the same either way.
