Circular Motion Equations Worksheet
Practice uniform circular motion with 10 problems covering centripetal acceleration, centripetal force, angular velocity, and period. Full solutions included.
View worksheet →Projectile Motion Equations Worksheet
Practice projectile motion with 10 problems covering horizontal launches, angled trajectories, time of flight, maximum height, and range. Resolving into x and y components is the key skill.
Equations you will need
| x = (u cos θ) t | Horizontal displacement |
| y = (u sin θ) t - ½gt² | Vertical displacement |
| vₓ = u cos θ | Horizontal velocity (constant) |
| vy = u sin θ - gt | Vertical velocity at time t |
| R = u² sin(2θ) / g | Range on level ground |
| t_flight = 2u sin θ / g | Total time of flight |
| H = u² sin²θ / (2g) | Maximum height |
Symbol key
| Symbol | Quantity | Unit |
|---|---|---|
| u | launch speed | m/s |
| θ | launch angle | degrees |
| g | gravitational acceleration | 9.8 m/s² |
| R | horizontal range | m |
| H | maximum height | m |
| t | time | s |
Practice problems
Attempt each problem on paper first, then click Show answer to check your working.
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A ball is launched horizontally at 20 m/s from a 45 m high cliff. Find time to hit ground. (g = 9.8 m/s²)
Show answer
45 = ½(9.8)t² → t = 3.03 s -
Same ball: how far from the base does it land?
Show answer
x = 20 × 3.03 = 60.6 m -
A projectile is launched at 30 m/s at 60° above horizontal. Find the time of flight.
Show answer
t = 2(30)sin60° / 9.8 = 5.30 s -
Same projectile: find the maximum height.
Show answer
H = (30² × sin²60°)/(2×9.8) = 34.4 m -
Same projectile: find the horizontal range.
Show answer
R = (30² × sin120°)/9.8 = 79.5 m -
A ball kicked at 25 m/s at 40° above horizontal. Find horizontal range.
Show answer
R = (25² × sin80°)/9.8 = 62.8 m -
A stone thrown horizontally at 15 m/s lands 30 m away. Find the launch height.
Show answer
t = 30/15 = 2 s; h = ½(9.8)(4) = 19.6 m -
A projectile launched at 20 m/s at angle θ has range 35 m. Find θ.
Show answer
sin(2θ) = (35)(9.8)/400 = 0.8575 → 2θ ≈ 59.0° → θ ≈ 29.5° -
Find the vertical velocity of a ball 2 s after launch at 40 m/s at 30°.
Show answer
vy = 40 sin30° - 9.8(2) = 20 - 19.6 = 0.4 m/s -
A projectile launched at 50 m/s at 45°. Find the speed when it lands at the same height.
Show answer
By symmetry, lands at 50 m/s (energy conserved)
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View worksheet →About this worksheet
This projectile motion equations worksheet covers the essential equations for mechanics at the A-Level / AP Physics 1 / IB level. Every problem has been written to mirror the style and difficulty of real exam questions, with full algebraic working shown in the solutions.
If you find these problems too straightforward, try the more advanced worksheets in the same topic listed above. If they feel too difficult, start by reviewing the equation definitions in the box at the top of this page and then return to question 1.