Physics · Motion In A Straight Line · NEET
g always points DOWN, so its sign depends only on the direction YOU chose as positive - not on how the ball was thrown. If you take up as positive, then a = -g for the whole flight (going up AND coming down). If you take down as positive, then a = +g throughout. The throw direction only changes the sign of the initial velocity u, never the sign of g. Pick one convention and keep it for the whole problem.
Use v^2 = u^2 + 2as, because it links initial velocity, final velocity and displacement without needing time. Example: thrown DOWN at u = 20 m/s, hits ground at v = 80 m/s, g = 10. Taking down positive: 80^2 = 20^2 + 2(10)s, so s = (6400 - 400)/20 = 300 m. The tower is 300 m tall. This is a real NEET 2020 question.
Yes. The one thrown up first rises, stops, then falls back past the bridge, so it spends extra time in the air. But here is the key NEET point: when both pass the bridge level again, the up-thrown ball has the SAME speed it was thrown with (just now pointing down). From that moment its motion is identical to the down-thrown ball. So the up-thrown one lands later, but both hit the water with the same final speed if thrown at the same speed.
'Dropped' or 'released' or 'let fall' means u = 0 - no initial push. 'Thrown down' means u is the given speed with the same sign as gravity (positive if down is positive). 'Thrown up' means u has the opposite sign to gravity. Reading this word correctly is half the problem; NEET often mixes 'dropped' and 'thrown' in the same paper to catch you.
Use signed displacement, not distance. Take down as positive from the launch point and plug the total time into s = ut + (1/2)at^2, where u is negative (it went up first). Example (NEET 2023): thrown up at 4 m/s from a bridge, hits water after 4 s, g = 10. s = -4(4) + (1/2)(10)(16) = -16 + 80 = 64 m below the launch point. So the bridge is 64 m above the water. The equation handles the up-then-down trip in one line.
For a ball thrown straight DOWN, distance and displacement are equal, so height = displacement works directly. But for a ball thrown UP, the total distance (up + down) is bigger than the height of the tower. The height above ground is the straight-line displacement from launch point to ground, which is exactly what s = ut + (1/2)at^2 gives you. Never add the upward path to the tower height.
A ball is thrown vertically downward with a velocity of 20 m/s from the top of a tower. It hits the ground after some time with a velocity of 80 m/s. The height of the tower is (g = 10 m/s^2):
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/s^2):
Try the real previous-year questions from this chapter — each with the answer and a full solution.
Use whatever value the question gives. NEET problems usually state g = 10 m/s^2 to keep the arithmetic clean; if nothing is stated, use 9.8 m/s^2. Always match the value in the question.
No. In free fall (ignoring air resistance) all objects have the same acceleration g regardless of mass. The ball's mass never appears in v = u + at, s = ut + (1/2)at^2, or v^2 = u^2 + 2as.
If time is not involved, use v^2 = u^2 + 2as (fastest for height from speeds). If time is given or asked, use s = ut + (1/2)at^2. To find final velocity from time, use v = u + at. Choose the one that avoids the unknown you do not need.
Exactly the same speed it was thrown with, but now directed downward. From that instant it behaves like a ball thrown straight down at that speed, so you can treat the rest of the fall as a simple downward throw.
By v^2 = u^2 + 2gh, the final speed depends only on u^2 (sign of u disappears when squared) and the drop h. Since both start with the same speed magnitude and fall the same height, both land with the same final speed - the up-thrown one just takes longer.
