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A ball thrown up from a balcony, a stone dropped from a bridge, two balls that meet in mid-air: these are some of the first mechanics questions you meet, and they keep appearing in JEE and board papers. Most errors here are not about formulas. They come from signs and from one stubborn idea about the top of the flight. This post fixes both, with three worked examples.
Why is the acceleration not zero at the top?
At the highest point the ball's velocity is zero for an instant. Many students conclude that its acceleration is zero too. It is not. The only force on the ball (ignoring air) is its weight, and that does not switch off at the top, so the acceleration stays g, downward, the whole time.
This is part of a wider pattern. In interviews and written tests, John Clement (1982) found that many physics students hold a stable alternative view of how force and acceleration are related, one that resists change even after instruction. A common form of it is the belief that a ball moving upward must have an upward force on it. Physics education researchers Trowbridge and McDermott (1981) also found that students who could recite a = Δv/Δt often could not apply it to real motions they watched. If you can say why the acceleration at the top is g, you have dealt with both problems.
Which sign convention should you use?
Choose upward as positive and put the origin at the launch point. Then a = −g throughout, and you never split the flight into an "up part" and a "down part" unless the question asks about them. NCERT's free-fall example uses the same choice: upward positive, so a = −g. NCERT also notes that g can be taken as constant only when the height is small compared with Earth's radius.
The equations you need are the constant-acceleration equations:
v = u − gt
s = ut − ½gt²
v² = u² − 2gs
Here s is the displacement from the launch point, not the distance travelled. A ball that lands below its launch point has negative s.
Worked example 1: thrown up from a height
A ball is thrown up at 15 m/s from a balcony 20 m above the ground. Take g = 10 m/s². When does it land, and how fast is it going?
Landing means s = −20 m. So −20 = 15t − 5t², which rearranges to t² − 3t − 4 = 0.
That factorises as (t − 4)(t + 1) = 0. Reject the negative root: t = 4 s.
Velocity at landing: v = 15 − 10 × 4 = −25 m/s, so 25 m/s downward.
Check with v² = u² − 2gs: 225 − 2 × 10 × (−20) = 625, so |v| = 25 m/s.
Along the way, the ball rises for 15/10 = 1.5 s and reaches 15²/(2 × 10) = 11.25 m above the balcony, 31.25 m above the ground.
Worked example 2: dropped from rest
A stone is dropped from a bridge and hits the water 3 s later. With u = 0, the drop is ½ × 10 × 3² = 45 m and the final speed is 30 m/s. Notice the pattern: in successive seconds from rest, a falling object covers 5 m, 15 m and 25 m, in the ratio 1 : 3 : 5. NCERT credits this to Galileo, the first to make quantitative studies of free fall.
How do you solve two-ball meeting problems?
Ball A is dropped from the top of an 80 m tower. At the same moment, ball B is thrown straight up from the ground at 40 m/s, directly below. When and where do they meet?
The slow way is to write a position equation for each ball and set them equal. The fast way is relative motion. Both balls have the same acceleration, g downward, so relative to A, ball B moves with no acceleration at all. B closes the 80 m gap at a steady 40 m/s.
Meeting time: 80 / 40 = 2 s.
Height of the meeting point: A has fallen ½ × 10 × 2² = 20 m, so they meet 60 m above the ground.
Check with B: 40 × 2 − 5 × 2² = 60 m.
B's velocity then: 40 − 20 = 20 m/s, still moving up.
The same idea works when one ball is thrown a few seconds after the other; just account for the head start before you use the closing speed.
What mistakes cost marks in free-fall questions?
Mixing sign conventions inside one problem, such as taking g positive and the throw also positive.
Using distance instead of displacement in s = ut − ½gt².
Rejecting the wrong root of a quadratic, or keeping both without asking what each means.
Assuming the acceleration is zero at the top, or that heavier objects fall faster. OpenStax states that, with air resistance negligible, all objects at a location fall with the same acceleration, independent of their mass.
Sketching the velocity-time graph helps: it is a straight line of slope −g crossing zero at the top. Our guide to reading physics graphs shows how to read areas and slopes from it.
A practice routine for vertical motion
Write "up = +, a = −g" at the top of every solution before anything else.
Solve five thrown-up problems where the ball lands below the launch point.
Do three meeting problems twice: once with two equations, once with relative motion.
For every answer, check one other equation (usually v² = u² − 2gs).
When a problem has several stages and you are not sure which equation to use next, the Fermi Problem Library has a step-by-step solver that explains the reasoning rather than printing an answer.
Tonight, solve worked example 1 again without looking, then change the balcony height and solve it once more.
