3.1 Motion of a projectile
- Syllabus
- 9231–2028–2029
- Topic
- 3.1
- Level
- A2
Learning objectives
3.1.1Projectile motion• model the motion of a projectile as a particle moving with constant acceleration and understand any limitations of the model Vector methods are not required3.1.2Projectile motion• use horizontal and vertical equations of motion to solve problems on the motion of projectiles, including finding the magnitude and direction of the velocity at a given time or position, the range on a horizontal plane and the greatest height reached3.1.3Projectile motion• derive and use the Cartesian equation of the trajectory of a projectile, including problems in which the initial speed and/or angle of projection may be unknown. Knowledge of the 'bounding parabola' for accessible points is not included.
The projectile model creates two synchronized scalar motions
| modelling assumption | mathematical consequence | limitation |
|---|---|---|
| projectile is a particle | size, shape and rotation are ignored | spin and dimensions cannot affect motion |
| air resistance and wind are ignored | no horizontal force after launch | real horizontal speed may decrease or drift |
| gravity is uniform and vertical | ax=0, ay=−g with constant g | suitable only over ordinary near-Earth distances |
| fixed ground frame | horizontal and vertical axes remain fixed | launch/landing geometry must be stated separately |
With initial speed $u$ at angle $\theta$ above the horizontal,u_x=u\cos\theta,\qquad u_y=u\sin\theta.The components share the same elapsed time, but horizontal velocity stays constant while vertical velocity changes under $-g$.
Gravity acts throughout ascent and descent. At the highest point only the vertical velocity is zero; the horizontal component usually remains non-zero. Vector methods are not required, so solve the two scalar directions and recombine only when needed.
Use one time in both directions, then recombine the velocity
From launch point $O$,x=(u\cos\theta)t,\qquad y=(u\sin\theta)t-\tfrac12gt^2,v_x=u\cos\theta,\qquad v_y=u\sin\theta-gt.At any time, speed is $\sqrt{v_x^2+v_y^2}$ and the direction satisfies $\tan\alpha=|v_y|/v_x$, with ascent/descent stated.
Let $u=20\text{ m s}^{-1}$, $\theta=30^\circ$ and $g=10\text{ m s}^{-2}$ on level ground. Then $v_x=10\sqrt3$ and $v_y=10-10t$. Greatest height occurs at $t=1$:H=10(1)-5(1)^2=5\text{ m}.Returning to $y=0$ gives $t=2$, soR=(10\sqrt3)(2)=20\sqrt3\text{ m}.
Immediately before landing, $(v_x,v_y)=(10\sqrt3,-10)$, so the speed is $20\text{ m s}^{-1}$ and the direction is $30^\circ$ below the horizontal. Equal launch and landing heights create this symmetric speed result.
Do not use 2usinθ/g or u2sin2θ/g when the landing height differs from the launch height. Instead solve the stated vertical position equation for time, discard negative times, and then use the horizontal motion.
Eliminate time to turn a flight into a Cartesian parabola
Fromx=u\cos\theta,t,\qquad y=u\sin\theta,t-\tfrac12gt^2,use $t=x/(u\cos\theta)$ to obtainy=x\tan\theta-\frac{gx^2}{2u^2\cos^2\theta}=xT-\frac{gx^2}{2u^2}(1+T^2),\qquad T=\tan\theta.Addtheinitialheightifthelaunchpointisnottheorigin.
If a known-speed projectile passes through $(X,Y)$, substitute the point to get a quadratic in $T$:\frac{gX^2}{2u^2}T^2-XT+\left(Y+\frac{gX^2}{2u^2}\right)=0.Each admissible real root gives a possible launch angle $\theta=\tan^{-1}T$; often these are low and high paths.
Iftheangleisknowninstead,rearrangethesamepointcondition:u^2=\frac{gX^2(1+T^2)}{2(XT-Y)},requiring $XT>Y$. Use intersections with ground, walls or targets only on the forward part of the flight, with $t=x/(u\cos\theta)\ge0$.
The trajectory equation assumes the same ideal model as the component equations. Retain every physically valid root and reject roots that violate speed, angle, time or geometry conditions. The bounding parabola for all accessible points is explicitly outside this syllabus.