Vectors and Projectile Motion

Starting point: what you already know

• Vectors in $\mathbf{i}$, $\mathbf{j}$ form, and $r\cos \theta \mathbf{i}+r\sin \theta \mathbf{j}$ for a vector of length $r$ at angle $\theta$.
• From Mathematics Advanced: velocity is the derivative of displacement, $v=\frac{dx}{dt}$, and acceleration is the derivative of velocity. Going backwards uses integration.
• Gravity gives every falling object an acceleration of about $g=9.8$ m/s² downward (often rounded to 10 in exam questions).
• Solving quadratics and finding the maximum of a parabola.

Setting up the vectors

Put the origin at the launch point, $\mathbf{i}$ horizontal and $\mathbf{j}$ up. The only force (ignoring air resistance) is gravity, so

$$\mathbf{a}(t)=-g\mathbf{j}$$

Launch with speed $u$ at angle $\theta$ to the horizontal:

$$\mathbf{v}(0)=u\cos \theta \mathbf{i}+u\sin \theta \mathbf{j}$$

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Integrate once for velocity and again for position (the constants come from the starting values):

Notice that the horizontal velocity $u\cos \theta$ never changes.

The three big questions

Use the free kick: $u=20$, $\theta =50^{\circ }$, $g=10$. Then $u\cos \theta \approx 12.86$ and $u\sin \theta \approx 15.32$.

1. Greatest height. At the top, the ball stops going up: vertical velocity is 0. $$15.32-10t=0\Rightarrow t\approx 1.53\text{ s}$$ $$y=15.32(1.53)-5(1.53{)}^2\approx 11.7\text{ m}$$

2. Time of flight. It lands when $y=0$: $$t(15.32-5t)=0\Rightarrow t\approx 3.06\text{ s}$$ This is twice the time to the top, as symmetry suggests.

3. Range. Horizontal distance at landing: $$x=12.86\times 3.06\approx 39.4\text{ m}$$

Quick summary

• Horizontal and vertical motion are separate. Gravity only affects the vertical part.
• $x=ut\cos \theta$ and $y=ut\sin \theta -\frac{1}{2}g{t}^2$.
• Greatest height: vertical velocity is 0. Flight time: $y=0$. Range: $x$ at landing.
• On flat ground the range is $\frac{u^2\sin 2\theta }{g}$, greatest at $45^{\circ }$.

Quiz