Maths › Mechanics › Projectiles
Projectiles
A projectile is two problems wearing one flight path: steady speed across, gravity down, sharing nothing but the clock. Split the launch velocity and each half is a chapter you already know.
Builds on Kinematics with constant acceleration and Triangles and the sine and cosine rules.
IN THIS TOPIC
- Resolve a launch velocity into horizontal and vertical components.
- Apply constant-velocity motion horizontally and suvat with g vertically, linked by time.
- Find time of flight, range, greatest height, and velocity at any instant.
WHAT YOU PROBABLY THINK
A projectile's horizontal speed gradually decreases during flight.
Two motions, one clock
Launching at speed u and angle θ gives components u cos θ across and u sin θ up. With air resistance modelled away, nothing horizontal ever changes the horizontal speed, and nothing vertical is anything but gravity. The two motions proceed independently, joined only by the shared time t.
The standard questions
WORKED EXAMPLE
Flight of a golf ball
A ball is struck at 20 m s⁻¹ at 30° above the horizontal from level ground. Find the time of flight and the range.
Components: across 20 cos 30° = 17.3, up 20 sin 30° = 10.
Vertical suvat to return to the ground: 0 = 10t − 4.9t², so t = 10/4.9 = 2.04 s.
Range: 17.3 × 2.04 = 35.3 m.
The check: greatest height 10²/(2 × 9.8) = 5.1 m at half the flight, 1.02 s, and the symmetry holds.
Greatest height is a vertical question: v = 0 upward, so h = (u sin θ)²/2g. Speed mid-flight reassembles the components with Pythagoras, and the direction of travel comes from tan of the ratio. A projectile launched horizontally is the same machine with u sin θ = 0: it simply starts at its own apex.
THE EXAM BIT
- First line every time: the two components, with cos on the horizontal.
- Time is the bridge: found from one direction, spent in the other.
- Landing on level ground means vertical displacement zero, not velocity zero.
- Answers to 2 or 3 significant figures, consistent with g = 9.8.
CHECK YOURSELF
A stone is thrown horizontally at 15 m s⁻¹ from a 19.6 m cliff. How long does it fall, and how far from the cliff base does it land?
Show a hint
The vertical motion is a straight drop from rest.
Show the answer
Vertically: 19.6 = 4.9t², so t = 2 s.
Horizontally: 15 × 2 = 30 m from the base.
Resolve once: u cos θ across for ever, u sin θ up into gravity's hands.
Solve whichever direction knows the answer, and carry t across to the other.
WORKBOOK
Printable practice for this topic: original exam-style questions with room to work, and a fully worked answer book. Free to use; please do not redistribute or sell.
CHECK YOUR PROGRESS
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- Resolve a launch velocity into horizontal and vertical components.
- Apply constant-velocity motion horizontally and suvat with g vertically, linked by time.
- Find time of flight, range, greatest height, and velocity at any instant.
No animated video for this topic yet; these notes stand alone.