Section 1 of 5Two Independent Motions
A projectile moves under gravity alone, and its horizontal and vertical motions do not affect each other.
Horizontal motion
- With no air resistance there is no horizontal force, so the horizontal velocity is constant.
Vertical motion
- The only force is weight, so the vertical acceleration is g = 9.81 m s⁻² downwards.
Linking them
- Time is the same for both motions, so it is the quantity that connects the two sets of equations.
Path
- Combining constant horizontal velocity with constant vertical acceleration gives a parabolic path.
Section 2 of 5Solving a Problem
1
Resolve the launch velocity into horizontal and vertical components if the launch is at an angle.
2
Write down the known values separately for horizontal and vertical motion.
3
Use the vertical SUVAT equations with a = −9.81 m s⁻² (or +9.81 if downwards is positive) to find the time.
4
Use the time in horizontal distance = horizontal velocity × time to find the range.
5
Combine the final components using Pythagoras to find the speed and the direction of motion at any point.
Section 3 of 5Key Points and Air Resistance
Key Points and Air Resistance
Top of flight
- At the highest point the vertical velocity is zero, but the horizontal velocity is unchanged.
Horizontal launch
- For a ball projected horizontally, the initial vertical velocity is zero.
Symmetry
- On level ground, the time up equals the time down and the launch and landing speeds are equal.
Air resistance
- In reality, drag reduces both the range and the maximum height, and the path is no longer a perfect parabola.
| Direction | Acceleration | Velocity |
|---|---|---|
| Horizontal | Zero | Constant |
| Vertical | 9.81 m s⁻² downwards | Changing steadily |
Section 4 of 5Worked Example
Question
- A ball is kicked horizontally at 12 m s⁻¹ from the top of a cliff 45 m high.
- Calculate the time to reach the ground and the horizontal distance travelled.
- Ignore air resistance.
- Vertical motionu = 0, a = 9.81 m s⁻², s = 45 m, using s = ut + ½at².
- 45 = ½ × 9.81 × t², so t² = 9.17 and t = 3.03 s.
- Horizontal distance = 12 × 3.03 = 36.3 m, and the vertical velocity on landing is 9.81 × 3.03 = 29.7 m s⁻¹.
Answer The ball takes 3.0 s and lands about 36 m from the base of the cliff.
Section 5 of 5Exam Checklist & Tips
3.2.5 – Key Points for the Exam
- Projectile An object moving through the air under the influence of gravity alone.
- Parabola The curved path followed by a projectile when air resistance is ignored.
- Range The horizontal distance travelled by a projectile.
- Time of flight The total time a projectile spends in the air.
- Free fall Motion under the sole influence of gravity.
Common Mistakes
- Using the total velocity in the horizontal distance equation instead of only the horizontal component.
- Assuming the horizontal velocity decreases as the projectile rises.
- Setting the vertical velocity to zero for the whole flight rather than only at the top.
- Forgetting that time of flight depends on the vertical motion only.
Physics 0625 & 9702 SUVAT Mastery
Graph Interpretation
Constant Acceleration Equations (SUVAT) & Slopes
v = u + at | s = ut + ½at2 | v2 = u2 + 2as | s = ½(u + v)t
- Displacement-Time Graph: Gradient = Velocity (curve = accelerating).
- Velocity-Time Graph: Gradient = Acceleration; Area under curve = Distance travelled / Displacement.
- Free Fall in vacuum:
a = g ≈ 9.81 m/s2downwards (air resistance negligible).