- Calculate speed, height and time for free fall and vertical throws
- Apply t₁ = v₀ / g, H = v₀² / (2g) and the symmetry of the motion
- Find the rise time and height from vy–t and y–t graphs
- Find the time of flight, height and range of horizontal and angled throws
A ball you throw up from a balcony first slows down, “freezes” in the air for an instant and then comes back faster and faster. Interestingly, going up and coming down take the same time, and the ball returns to the launch level with the speed it was thrown at. In this lesson we study motion under gravity alone: free fall, vertical throws up and down, horizontal throws and throws at an angle. In the 2026 DİM papers the rise time of such a body had to be read from its vy–t and y–t graphs.
Free fall
Motion of a body under gravity alone (air resistance is neglected). All bodies, whatever their mass, move with the same free-fall acceleration: g ≈ 9.8 m/s² at the Earth’s surface, and g = 10 m/s² is used in problems. It always points vertically down.
Free fall is uniformly accelerated motion that starts from rest, so its formulas are the usual ones with v₀ = 0 and a = g.
- vvelocity at time t, in m/s
- hheight fallen in time t, in m
- gfree-fall acceleration, 10 m/s²
The formulas of free fall (v₀ = 0).
A stone is dropped from rest from a height of 45 m. Find the fall time, its speed when it hits the ground and the distance it falls in the last second.
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v = g · t = 10 · 3 = 30 m/s (check: √(2 · 10 · 45) = 30).
In the first 2 s: 10 · 2² / 2 = 20 m, so in the last second 45 − 20 = 25 m.
A body thrown vertically up
Take the Oy axis pointing up with its origin at the launch point. The velocity points up and the acceleration down, so vy drops by g every second. At the top vy = 0 — but the acceleration is still g! After that the body falls freely. Rise and fall are symmetric: at any height the speed on the way up equals the speed on the way down.
- v₀initial speed, in m/s
- vyvelocity projection: positive going up, negative coming down
- yheight above the launch point, in m
Equations of motion for a body thrown straight up.
- t₁rise time (from vy = 0)
- Hmaximum height
- Ttime to return to the launch level, T = 2t₁
Symmetry: the body comes back with speed v₀.
A ball is thrown straight up from the ground at 25 m/s. Find the rise time, the maximum height, the time of flight, and the velocity and height at t = 1 s and t = 4 s.
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t = 1 s: vy = 25 − 10 = 15 m/s (up), y = 25 − 5 = 20 m.
t = 4 s: vy = 25 − 40 = −15 m/s (down), y = 100 − 80 = 20 m.
The ball passes the height of 20 m twice, at 1 s and at 4 s, both times at 15 m/s: that is the symmetry (1 s and 4 s are equally far from t₁ = 2.5 s).
Match them. While a body thrown straight up falls from the top back to the ground (no air resistance), its: 1. increases; 2. decreases; 3. stays the same.
a. height above the ground; b. speed; c. acceleration; d. kinetic energy; e. total mechanical energy.
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Answer: 1 – b, d; 2 – a; 3 – c, e.
A body thrown vertically down
Here both the velocity and the acceleration point down, so it is convenient to point the Oy axis down. The speed starts at v₀ and grows by g every second.
- v₀initial downward speed, in m/s
- hheight fallen, in m
A stone is thrown straight down at 5 m/s from a height of 30 m. How long does it take to land, and at what speed? How long would it take if it were simply dropped?
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v = 5 + 10 · 2 = 25 m/s; check: v² = 5² + 2 · 10 · 30 = 625 → v = 25 m/s.
Dropped: t = √(2 · 30 / 10) = √6 ≈ 2.45 s, a little longer.
v_y–t and y–t graphs
- vy–t is a straight line with slope −g: it starts at v₀, crosses the time axis at t₁ and returns to the launch level at −v₀. The area of the triangle above the axis is H.
- y–t is a downward-opening parabola with its top at (t₁, H); it meets the time axis at 0 and T = 2t₁ and is symmetric about t₁.
- ay–t is a line parallel to the time axis at −g: the acceleration never changes.
The vy–t graph of a body thrown straight up from the ground is a straight line. The initial velocity is not marked, but the line passes through (7 s, −30 m/s). Calculate the time it takes the body to reach its maximum height (g = 10 m/s², no air resistance).
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Method II. After the top the body falls freely and reaches 30 m/s in 30 / 10 = 3 s. So t₁ = 7 − 3 = 4 s.
In a written answer show the formula, the substitution and the unit. Bonus: H = 40² / 20 = 80 m.
The y–t graph of a body thrown straight up from the ground is part of a parabola through the origin and the point (2 s, 40 m). Find the time the body takes to reach its maximum height (g = 10 m/s²).
A) 2 s B) 3 s C) 4 s D) 2.5 s E) 6 s
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t₁ = v₀ / g = 30 / 10 = 3 s (B).
Traps: 2 s treats the given point as the top; 6 s is the whole flight (2t₁).
Horizontal throws and throws at an angle
The key idea: the motions are independent. No force acts horizontally, so the body moves uniformly in that direction. Vertically it is ordinary free fall or a vertical throw. That is why a ball thrown horizontally and a ball simply dropped from the same height at the same moment land together.
- v₀horizontal launch speed, in m/s
- hlaunch height, in m
- Rrange, in m
Horizontal throw: the time of flight does not depend on v₀. The landing speed is v = √(v₀² + (gt)²).
Murad throws a stone horizontally at 15 m/s from a 20 m high cliff. Find the time of flight, the range, the landing speed and the angle the velocity makes with the horizontal.
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vy = g · t = 20 m/s, vₓ = 15 m/s → v = √(15² + 20²) = 25 m/s.
tan β = vy / vₓ = 20 / 15 = 4/3 → β ≈ 53°.
For a body thrown at an angle α to the horizontal we split the velocity into two components: v₀ₓ = v₀ · cos α (constant) and v₀y = v₀ · sin α (changes as in a vertical throw). At the top the speed is not zero — it equals v₀ · cos α.
- αlaunch angle (between the velocity and the horizontal)
- T, Htime of flight and maximum height
- Rrange (landing at the launch level)
The range is greatest at 45°; angles that add up to 90° (30° and 60°) give the same range.
A ball is kicked at 20 m/s at 30° to the horizontal. Find the time of flight, the maximum height and the range. What would change at 60°?
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T = 2 · 10 / 10 = 2 s; H = 10² / 20 = 5 m; R = 17.3 · 2 ≈ 34.6 m (= 20² · sin 60° / 10).
At 60°: R is the same (34.6 m), but H = 15 m and T ≈ 3.46 s — the ball flies higher and longer.
- 1.A freely falling body covers m in 4 s.
- 2.A body thrown up at 30 m/s rises for s.
- 3.A body thrown up at 30 m/s reaches a maximum height of m.
- 4.A body thrown horizontally from a height of 80 m flies for s.
Key points
- Free fall is uniformly accelerated motion with v₀ = 0 and a = g: v = gt, h = gt²/2; with g = 10 m/s² it covers 5, 15, 25, … m in successive seconds.
- For an upward throw t₁ = v₀ / g, H = v₀² / (2g), the return time is 2t₁ and the return speed is v₀.
- At the top vy = 0, but the acceleration is always g, directed down.
- vy–t is a straight line with slope −g, and y–t is a parabola symmetric about t₁.
- In horizontal and angled throws the motions are independent: vₓ is constant, and vertically the body falls freely.
Check yourself
12 questions. Every correct answer earns XP.