Lesson summary
2.1.9 Projectile Motion — AS PHYSICS 9702 notes: Analyse projectile motion by treating the horizontal and vertical components independently.A projectile moves under the influence of gravity alone — with air resistance negligible, for example a thrown ball or a kicked football.The…
- The motion has a vertical component and a horizontal component
- After release the only force is gravity
- Time of flight
- Maximum height
- Resolve the launch velocity u at angle θ above the horizontal
- The speed at any moment is √(vx2 + vy2)
Section 1 of 11Projectile motion
Projectile motion
A projectile moves under the influence of gravity alone
- A projectile moves under the influence of gravity alone with air resistance negligible, for example a thrown ball or a kicked football.
The path is a parabola.
The motion has a vertical component and a horizontal component
- The motion has a vertical component and a horizontal component and they are completely independent, so they are evaluated separately.
After release the only force is gravity
- After release the only force is gravity so there is no horizontal acceleration.
Section 2 of 11Quantities to find
Time of flight
Time of flight how long the projectile is in the air.
Maximum height
Maximum height the height at which the projectile is momentarily at rest vertically.
Range
Range the horizontal distance travelled by the projectile.
Time is the quantity shared by both directions.
Section 3 of 11Horizontal and vertical equations
1Resolve the launch velocity u at angle θ above the horizontal
- Resolve the launch velocity u at angle θ above the horizontal ux = u cos θ and uy = u sin θ.
2Horizontal
- Horizontal acceleration is zero, so the velocity is constant and x = uxt.
3Vertical
- Vertical constant acceleration −g (up positive), so vy = uy − gt, y = uyt − ½gt2 and vy2 = uy2 − 2gy.
4The speed at any moment is √(vx2 + vy2)
- The speed at any moment is √(vx2 + vy2) at an angle tan-1(vy ÷ vx) to the horizontal.
Section 4 of 11Vertical projection
1
The object moves up and down only.
2At the top v = 0
- At the top v = 0 so the maximum height is found from v2 = u2 + 2as with a = −g.
3
The time up equals the time down to the same level.
4Falling from rest through height s
- Falling from rest through height s s = ½gt2.
Section 5 of 11Horizontal projection
Horizontal projection
An object launched horizontally from height h has uy = 0
- An object launched horizontally from height h has uy = 0 the time to fall comes from h = ½gt2, so t = √(2h ÷ g).
The time to fall does not depend on the horizontal speed.
Range R = uxt.
A ball rolled off a table and a ball dropped from the edge land at the same moment
- their vertical motions are identical.
The path is half a parabola.
Section 6 of 11Projection at an angle (level ground)
Time to the top
- Time to the top t = u sin θ ÷ g, when vy = 0.
Maximum height
- H = u2 sin2θ ÷ 2g.
Total time of flight
- T = 2u sin θ ÷ g.
Range
- R = u2 sin 2θ ÷ g, greatest at θ = 45°.
Complementary angles give the same range
- Complementary angles give the same range such as 30° and 60°.
At the top the velocity is horizontal, u cos θ
- At the top the velocity is horizontal, u cos θ and the acceleration is still g downward.
The speed on landing at the same level equals the launch speed.
Section 7 of 11Solving a projectile problem
1
Resolve the launch velocity into horizontal and vertical components.
2
Make two columns of s, u, v, a and t: horizontal with a = 0 and vertical with a = −g.
3
Use the direction that gives enough information to find the time t.
4
Substitute t into the other direction.
5
For landing at a different height solve y = uyt − ½gt2 as a quadratic and accept the physically sensible root.
6
Find the landing angle from tan-1(|vy| ÷ vx).
Section 8 of 11Air resistance
1
Air resistance reduces both range and maximum height.
2The path becomes asymmetrical
- The path becomes asymmetrical steeper on the way down.
3
The best angle for range falls below 45°.
4Questions normally state
- Questions normally state that air resistance can be neglected.
Section 9 of 11More from the specification
More from the specification
Kinematics curves and light gates
- Kinematics curves and light gates plotting displacement-time graphs from light gates data validates independent horizontal and vertical parabolic flight paths.
Displacement-Time Graph Interpretation
- On a displacement-time graph, the slope or gradient at any point represents instantaneous velocity,
- while a straight line indicates constant velocity motion.
Velocity-Time Graph Characteristics
- On a velocity-time graph, the gradient represents instantaneous acceleration,
- and the area enclosed between the line and the time axis represents total displacement.
Projectile Motion Kinematic Equations
- For a projectile launched with initial velocity u at angle θ, horizontal displacement is x = (u cos θ)t,
- while vertical motion follows kinematic equations such as y = (u sin θ)t - ½gt².
Light gate
- A sensor that starts or stops a timer when a beam is interrupted.
Scalar quantity
- A quantity with magnitude only, such as potential
Electromagnet
- Magnet made by current through a coil around a magnetic core.
Velocity-time graph
- A graph of velocity against time whose gradient is acceleration and area is displacement.
Initial velocity
- Velocity at the start, u.
Displacement-time graph
- A graph of displacement against time whose gradient is velocity.
Kinematic equations
- Equations relating displacement, velocity, acceleration and time for constant acceleration.
Final velocity
- Velocity at the end, v.
Acceleration-time graph
- A graph of acceleration against time whose area is the change in velocity.
Vector quantity
- A quantity with both magnitude and direction.
Displacement
- Change in position.
Gradient
- Slope of a graph line.
Section 10 of 11Key terms
| Term | Meaning |
|---|---|
| Projectile | An object moving under gravity alone. |
| Time of flight | The time a projectile is in the air. |
| Range | The horizontal distance travelled by a projectile. |
| Maximum height | The greatest height reached, where the vertical velocity is zero. |
| Component | The part of a vector along one of two perpendicular directions. |
| Trajectory | The path followed by a projectile, a parabola. |
| Independent motions | The horizontal and vertical motions do not affect each other. |
| Parabola | The curved shape of a projectile's path. |
Section 11 of 11Exam technique
1
Using the horizontal speed in the vertical equations, or applying g to the horizontal motion.
2
Thinking the velocity is zero at the top — only the vertical component is zero.
3
Assuming a faster horizontal launch makes the object take longer to fall.
4
Applying R = u² sin 2θ ÷ g when the landing height differs from the launch height.
5
Forgetting that the rising object has a negative acceleration −g.
Marked practiceQuizzes for this lesson
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).