A.1.8—Projectile components
- Syllabus
- First assessment 2025
- Objective
- —
- Level
- HL
Separate the axes
With negligible fluid resistance, projectile motion is independent horizontal and vertical motion. Resolve the launch velocity:
ux=ucosθ,uy=usinθ
Horizontal motion
There is no horizontal acceleration in the ideal model, so vx=ux and x=uxt. Use the horizontal displacement to find time or horizontal speed.
Vertical motion
Use one-dimensional constant-acceleration equations vertically, usually with ay=−g if upward is positive. The horizontal and vertical equations share the same time t.
Worked example from local Question Bank row 31723
A tennis ball travels 11.9m horizontally after launch at 64.0ms−1 and 7∘ to the horizontal.
ux=64.0cos7∘=63.52ms−1
t=uxx=63.5211.9=0.187s
The same 0.187s must then be used in the vertical equation.
Common trap
Do not use the launch speed as the horizontal speed. Resolve it first, and do not use the horizontal time independently of the vertical motion.
[ 2 ]
The ball leaves the ground at an angle of 22∘. The horizontal distance from the initial position of the edge of the ball to the wall is 11 m . Calculate the time taken for the ball to reach the wall.
Describe motion
Position locates the object, velocity is the rate of change of position, and acceleration is the rate of change of velocity. Distance and speed are scalar; displacement and velocity are directed quantities.
Use the right model
For uniform acceleration:
v=u+at,\quad s=ut+rac12at^2,\quad v^2=u^2+2as
For projectiles without drag, solve horizontal and vertical components with a shared time. Do not use these equations when acceleration is non-uniform.
Check the boundary
Ask whether the quantity is average or instantaneous, whether the route or endpoints matter, whether acceleration is constant, and whether a neglected force such as drag changes the model.