SPEC 7.3 • KINEMATICS
SUVAT Equations: Vertical Motion
Motion under gravity — objects projected upward, dropped, and thrown
Motion Under Gravity
Key Constant
Acceleration due to gravity:g = 9.8 m s⁻²
When an object moves vertically under gravity:
Gravity always acts downward — toward the centre of the Earth
The acceleration is constant (assuming no air resistance)
Objects speed up when falling, slow down when rising
The SUVAT equations apply with a = ±g
Modelling Assumption
We treat objects as particles and ignore air resistance. This is a good approximation for dense, small objects over short distances.
Sign Conventions
Choosing a consistent sign convention is essential for vertical motion problems.
Convention 1: Upward Positive
+ve direction
a = −g = −9.8 m s⁻²
Upward velocity: positive
Downward velocity: negative
Height above start: positive
Most common convention
Convention 2: Downward Positive
+ve direction
a = +g = +9.8 m s⁻²
Downward velocity: positive
Upward velocity: negative
Depth below start: positive
Useful for dropping problems
Critical Rule
Choose ONE convention and stick with it throughout the entire problem! Mixing conventions is the most common source of errors.
SUVAT Equations with Gravity
Using the upward positive convention (a = −g):
v = u − gt
s = ut − ½gt²
s = ½(u + v)t
v² = u² − 2gs
s = vt + ½gt²
Remember
These are the same SUVAT equations from Lesson 7.03, but with a replaced by −g (for upward positive). The physics doesn't change — only the direction of acceleration is specified.
Exam Tip
Always check what value of g the question specifies. Use g = 9.8 m s⁻² unless told otherwise.
Objects Projected Upward
When an object is thrown vertically upward from ground level:
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Key Results
QuantityFormulaNoteTime to max heightt = u/gWhen v = 0Maximum heighth = u²/(2g)Using v² = u² − 2gsTotal flight timeT = 2u/gTime up = Time downReturn velocityv = −uSame speed, opposite direction
01Worked Example
A ball is thrown vertically upward from ground level with initial velocity 19.6 m s⁻¹. Taking g = 9.8 m s⁻², find:
(a) the maximum height reached
(b) the time taken to reach maximum height
Taking upward as positive: u = 19.6 m s⁻¹, a = −9.8 m s⁻², v = 0 at max height
(a) Maximum height:
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1. Use v² = u² + 2as with v = 0
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2. 0 = 384.16 + 2(−9.8)s
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3. 0 = 384.16 − 19.6s → s = 19.6 m
(b) Time to max height:
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1. Use v = u + at with v = 0
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2. 0 = 19.6 + (−9.8)t
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3. 9.8t = 19.6 → t = 2 s
Worked Example02
A stone is projected vertically upward with velocity 24.5 m s⁻¹. Taking g = 9.8 m s⁻², find the total time of flight (time until it returns to its starting point).
Method 1: Using s = 0 for return
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1. When stone returns: s = 0, u = 24.5 m s⁻¹, a = −9.8 m s⁻²
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2. s = ut + ½at² → 0 = 24.5t − 4.9t²
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3. 0 = t(24.5 − 4.9t)
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4. t = 0 (start) or t = 24.5/4.9 = 5 s
Method 2: Using T = 2u/g
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1. T = 2 × 24.5/9.8 = 49/9.8 = 5 s ✓
Objects Dropped from Rest
When an object is dropped (released from rest):
Initial velocity u = 0
Object accelerates downward at g
Taking downward as positive simplifies the equations
Simplified Equations (u = 0, downward +ve)
v = gts = ½gt²v² = 2gs
Key Insight
For dropped objects, the velocity increases linearly with time, while the distance fallen increases with t². This means an object falls further in each successive second.
Time (s)Velocity (m s⁻¹)Distance fallen (m)
19.84.9
219.619.6
329.444.1
Using g = 9.8 m s⁻²
03Worked Example
An object is dropped from the top of a 44.1 m tower. Taking g = 9.8 m s⁻², find:
(a) the time taken to hit the ground
(b) the velocity on impact
Taking downward as positive: u = 0, s = 44.1 m, a = 9.8 m s⁻²
(a) Time to hit ground:
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1. Use s = ut + ½at² with u = 0
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2. 44.1 = 0 + ½(9.8)t²
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3. 44.1 = 4.9t² → t² = 9 → t = 3 s
(b) Impact velocity:
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1. Use v² = u² + 2as
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2. v² = 0 + 2(9.8)(44.1) = 864.36
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3. v = 29.4 m s⁻¹
Projection from Heights
When an object is projected from an elevated position (cliff, building, etc.), we need to consider:
Thrown Upward from Height
Object rises first, then falls
May pass starting level on way down
Final displacement is negative (below start)
Often gives quadratic with two solutions
Thrown Downward from Height
Initial velocity same direction as g
Object speeds up throughout
Takes less time than dropping
Higher impact velocity
Sign Convention Reminder
If using upward positive and the object lands below its starting point, the final displacement s is negative. For example, if thrown from a 40 m cliff and lands at ground level, s = −40 m.
04Worked Example
A stone is thrown vertically upward at 14.7 m s⁻¹ from the top of a 39.2 m cliff. Taking g = 9.8 m s⁻², find:
(a) the time taken to hit the ground
(b) the velocity on impact
Taking upward as positive: u = 14.7 m s⁻¹, a = −9.8 m s⁻², s = −39.2 m (lands below start)
(a) Time to hit ground:
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1. s = ut + ½at² → −39.2 = 14.7t − 4.9t²
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2. 4.9t² − 14.7t − 39.2 = 0 → t² − 3t − 8 = 0
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3. t = (3 ± √(9 + 32))/2 = (3 ± √41)/2
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4. t = (3 + 6.4)/2 = 4.7 s (reject negative)
(b) Impact velocity:
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1. v² = u² + 2as = 225 + 2(−10)(−40) = 225 + 800 = 1025
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2. v = −√1025 = −32.0 m s⁻¹ (negative = downward)
Worked Example05
An object is thrown vertically downward at 4.9 m s⁻¹ from the top of a 78.4 m building. Taking g = 9.8 m s⁻², find the velocity on impact with the ground.
Taking downward as positive: u = 4.9 m s⁻¹, s = 78.4 m, a = 9.8 m s⁻²
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1. Use v² = u² + 2as
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2. v² = 24.01 + 2(9.8)(78.4)
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3. v² = 24.01 + 1536.64 = 1560.65
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4. v = 39.5 m s⁻¹
Compare: If dropped (u = 0): v² = 1536.64, v = 39.2 m s⁻¹
Throwing downward adds only 0.3 m s⁻¹ — the 80 m fall dominates!
Two-Solution Problems
When finding the time at which an object reaches a certain height, the quadratic equation often gives two valid solutions:
Interpretation
t₁ = time when object passes height h on the way UP
t₂ = time when object passes height h on the way DOWN
Both solutions are physically meaningful!
The question context tells you which one to use
Quick Check
For an object projected upward: t₁ + t₂ = 2u/g (total flight time). The two times are symmetric about the time of maximum height.
Worked Example06
A ball is thrown vertically upward from ground level at 24.5 m s⁻¹. Taking g = 9.8 m s⁻², find the two times when the ball is 19.6 m above the ground.
Taking upward as positive: u = 24.5 m s⁻¹, s = 19.6 m, a = −9.8 m s⁻²
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1. s = ut + ½at² → 19.6 = 24.5t − 4.9t²
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2. 4.9t² − 24.5t + 19.6 = 0
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3. t² − 5t + 4 = 0
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4. (t − 1)(t − 4) = 0
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5.t = 1 s (going up) and t = 4 s (coming down)
Check: Max height at t = 24.5/9.8 = 2.5 s. Times 1 s and 4 s are symmetric about 2.5 s ✓
Worked Example07
A ball A is dropped from rest from a height of 78.4 m. At the same instant, ball B is projected vertically upward from ground level at 29.4 m s⁻¹. Taking g = 9.8 m s⁻², find when and where the balls meet.
Taking upward as positive, origin at ground level:
Ball A (dropped from 80 m):
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sA = 78.4 + 0 − 4.9t² = 78.4 − 4.9t²
Ball B (projected up from ground):
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sB = 0 + 29.4t − 4.9t² = 29.4t − 4.9t²
When they meet: sA = sB
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1. 78.4 − 4.9t² = 29.4t − 4.9t²
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2. 78.4 = 29.4t → t = 78.4/29.4 ≈ 2.67 s
Height where they meet:
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3. sB = 29.4(2.67) − 4.9(2.67)² = 78.5 − 34.9 = 43.6 m
Practice
A stone is thrown vertically upward from ground level at 14 m s⁻¹. Taking g = 9.8 m s⁻², find:
(a) the maximum height reached
(b) the total time of flight
(a) Maximum height: Using v² = u² − 2gs with v = 0
0 = 196 − 19.6s → s = 196/19.6 = 10 m
(b) Total flight time: T = 2u/g = 28/9.8 = 2.86 s
Practice
A ball is dropped from the top of a building and hits the ground after 3 seconds. Taking g = 9.8 m s⁻², find:
(a) the height of the building
(b) the velocity of the ball just before impact
(a) Height: s = ½gt² = ½(9.8)(9) = 44.1 m
(b) Impact velocity: v = gt = 9.8 × 3 = 29.4 m s⁻¹
Alternative for (b): v² = 2gs = 2(9.8)(44.1) = 864.36, so v = 29.4 m s⁻¹ ✓
Summary
Key Results (Upward Positive)
SituationKey Formula
Time to max heightt = u/g
Maximum heighth = u²/(2g)
Total flight time (return to start)T = 2u/g
Dropped object: velocity after time tv = gt
Dropped object: distance fallens = ½gt²
Exam Essentials
State your sign convention at the start of every solution
At maximum height: v = 0
Two solutions from a quadratic often both have physical meaning
Check: Does your answer make sense? (positive time, reasonable values)
Use g = 9.8 m s⁻² unless told otherwise
Next: Motion Under Gravity