Equations of Motion Under Gravity
When an object moves vertically under gravity alone (ignoring air resistance), it is in free fall. The only acceleration acting is gravitational acceleration g = 9.8 m·s⁻² directed downward.
Using upward as positive, we substitute a = −9.8 m·s⁻² into the standard equations of motion:
vf = vi + aΔt
Δy = viΔt + ½aΔt²
vf² = vi² + 2aΔy
| Symbol | Meaning | Unit |
|---|
| vi | Initial velocity | m·s⁻¹ |
| vf | Final velocity | m·s⁻¹ |
| Δy | Displacement (vertical) | m |
| a | Acceleration (= −9.8 m·s⁻²) | m·s⁻² |
| Δt | Time interval | s |
Sign Convention
The most common choice in vertical projectile problems is:
- Upward = positive direction
- Downward = negative direction
- Therefore a = −9.8 m·s⁻² (acceleration is always downward)
You may choose down as positive — but be consistent throughout. In all CAPS examples, up is taken as positive unless stated otherwise.
Object Thrown Upward
When an object is thrown upward with initial velocity vi > 0:
- It decelerates at 9.8 m·s⁻² (gravity acts opposite to motion)
- At the highest point, vf = 0 (momentarily at rest)
- It then falls back down, accelerating at 9.8 m·s⁻² downward
- If it returns to the same level, its speed at that point equals vi (by symmetry)
Maximum height: vf² = vi² + 2aΔy → 0 = vi² − 2(9.8)Δymax
Δymax = vi² / (2 × 9.8)
Free Fall from Rest
An object released from rest has vi = 0. Taking downward as positive (a = +9.8 m·s⁻²):
Δy = ½gΔt² (distance fallen from rest)
vf = gΔt (speed after time t)
Or using up as positive with a = −9.8 m·s⁻²: Δy is negative (displacement is downward).
Objects Dropped from a Height / Thrown Downward
If thrown downward: vi is negative (using up positive convention). The object accelerates further downward.
Strategy for all problems:
1. Define a positive direction (usually up).
2. Write down known values — assign correct signs.
3. Identify the unknown.
4. Choose the equation containing only one unknown.
5. Substitute and solve. Check sign of answer for direction.
Time of Flight and Velocity at Impact
For an object thrown upward from the ground and landing at the same level:
Total time of flight: Δttotal = 2vi / 9.8
Speed at impact = vi (by symmetry, speed going down = speed going up at same height)
Velocity-Time Graph for Vertical Projectile
The v-t graph for an object in free fall is always a straight line with gradient = a = −9.8 m·s⁻².
- Object thrown up: starts at +vi, decreases linearly to 0 at max height, continues to negative values as it falls
- When v = 0: maximum height reached
- The area under the v-t graph gives the displacement (Δy)
- The gradient at any point = −9.8 m·s⁻² (constant throughout)
The gradient of a v-t graph = acceleration. For free fall, gradient = −9.8 m·s⁻² (taking up as positive).
Worked Example 1 — Ball Thrown Upward
Problem: A ball is thrown upward from the ground at 20 m·s⁻¹. Calculate: (a) maximum height, (b) time to reach maximum height, (c) total time of flight, (d) speed when it returns to the ground.
Given: vi = +20 m·s⁻¹, a = −9.8 m·s⁻², at max height vf = 0
(a) vf² = vi² + 2aΔy → 0 = 400 + 2(−9.8)Δy → Δy = 400/19.6 = 20.4 m
(b) vf = vi + aΔt → 0 = 20 + (−9.8)Δt → Δt = 20/9.8 = 2.04 s
(c) By symmetry: total time = 2 × 2.04 = 4.08 s
(d) By symmetry: speed at ground = 20 m·s⁻¹ (downward, so vf = −20 m·s⁻¹)
Worked Example 2 — Dropped from Height
Problem: A stone is dropped from a bridge 45 m above a river. How long does it take to hit the water and what is its speed at impact?
Convention: Down = positive, so a = +9.8 m·s⁻², vi = 0, Δy = +45 m
Time: Δy = viΔt + ½aΔt² → 45 = 0 + ½(9.8)Δt² → Δt² = 9.18 → Δt = 3.03 s
Speed: vf = vi + aΔt = 0 + 9.8 × 3.03 = 29.7 m·s⁻¹
IEB Extension — Non-uniform Fields and Multi-stage Problems
IEB papers often include multi-stage problems where an object is launched, reaches a height, then is acted on by a second force or lands on a different level. Treat each stage separately:
- Use final conditions of one stage as initial conditions for the next
- Staggered launches: two objects launched at different times — set up equations for both and solve simultaneously
- Objects landing above/below launch point: Δy ≠ 0 when using quadratic equation
Δy = viΔt + ½aΔt² → quadratic in Δt → use quadratic formula
For staggered launches: if object A is launched t₀ seconds before object B, object A has been in flight for (t + t₀) while object B has been in flight for t seconds at the same clock time. Set their positions equal to find when/where they meet.