Treat Up and Down as One Motion with Constant Acceleration g
Anything moving vertically under gravity alone has the constant acceleration g downward, whether it is rising, at the top or falling. Choose a positive direction once, keep g's sign consistent, and every vertical problem is one of the constant-acceleration equations.
Motion and force questions become simpler when direction, starting condition, time and the acting forces are marked first.
Successive distances and times
A body dropped from rest covers distances h = (1/2) g t², so the time to fall the first n metres is proportional to sqrt(n), and the time to fall the nth successive metre is proportional to sqrt(n) - sqrt(n - 1). The times are not equal, and they decrease as the speed builds up. For two balls dropped from different heights that land together, 2 s apart, if the first takes 5 s (125 m) the second falls for 3 s (45 m), so the heights differ by 80 m.
Up-and-down symmetry
A body thrown up passes each height twice, once going up and once coming down, at the same speed.
the gap between passes is 2(u/g - t1).
A stone that passes A at speed v and a point B, 3 m higher, at v/2 has v² = 8g, so it rises (v/2)²/(2g) = 1 m more above B.
Balloons and relative motion in free fall
A stone released from a rising balloon starts with the balloon's upward velocity, so it first keeps moving up and only then falls. For a balloon rising at 2 m/s² for 10 s, the speed is 20 m/s and the height is 100 m; the stone rises 20 m more and then falls 120 m, so it covers 140 m in reaching the ground. Two bodies both under gravity have zero relative acceleration, so their relative velocity stays constant: a ball thrown up at u and another dropped at the same instant stay at a constant relative speed u.
Reading a height-time graph
A height-time plot for a ball thrown upward is a downward parabola, symmetric about its peak. With g = 7.5 m/s², read the time to the peak and use h = (1/2) g t² to find the maximum height, or use the symmetric times to apply h = (1/2) g t1 t2 at any height.
OlympIQ exam lens
- ✓Choose an upward-positive or downward-positive sign convention and keep g's sign consistent
- ✓Time to fall the nth successive metre is proportional to sqrt(n) - sqrt(n - 1)
- ✓Up and down times through the same height satisfy t1 + t2 = 2u/g and h = (1/2) g t1 t2
- ✓Up, down and free-fall times from a tower are linked by t = sqrt(t1 t2)
- ✓An object released from a moving platform keeps the platform's velocity at the instant of release
Do not assume a stone dropped from a rising balloon starts from rest. It inherits the balloon's upward velocity, so it first rises, and the distance it travels is the upward leg plus the longer fall to the ground.
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