A ball thrown through the air moves sideways and up-and-down at the same time. Parametric equations let us handle the two motions separately, after splitting the launch velocity into components.
Key idea
For an object launched at speed $v_0$ ft/s and angle $\theta$ from height $h$ ft (ignoring air resistance): $$x = (v_0\cos\theta)\,t, \qquad y = h + (v_0\sin\theta)\,t - 16t^2.$$ In meters, use $-4.9t^2$. It is highest when $t = \dfrac{v_0\sin\theta}{32}$ and lands when $y = 0$.
Why it works
The launch velocity is a vector with horizontal part $v_0\cos\theta$ and vertical part $v_0\sin\theta$. Nothing pushes the ball sideways, so the horizontal part stays constant and $x$ grows steadily. Gravity pulls down with acceleration $32$ ft/s², which subtracts $\frac{1}{2} \cdot 32 \cdot t^2 = 16t^2$ from the height. The height is a downward parabola in $t$, so its top is at the vertex, $t = -\frac{b}{2a} = \frac{v_0\sin\theta}{32}$.
Worked examples
Worked example: From the ground
A ball is kicked from the ground at $64$ ft/s at $30^\circ$. Find its greatest height, when it lands and how far it goes.
- Components: $64\cos 30^\circ = 32\sqrt{3}$ and $64\sin 30^\circ = 32$. So $x = 32\sqrt{3}\,t$ and $y = 32t - 16t^2$.
- Top at $t = \frac{32}{32} = 1$ s, where $y = 32 - 16 = 16$ ft.
- Lands when $16t(2 - t) = 0$: $t = 2$ s. Then $x = 64\sqrt{3} \approx 110.9$ ft.
Answer: $16$ ft high, lands at $2$ s, about $110.9$ ft away
Worked example: From a height
A ball is thrown from $6$ ft up at $80\sqrt{2}$ ft/s at $45^\circ$. How high is it after $2$ seconds?
- $80\sqrt{2}\sin 45^\circ = 80\sqrt{2} \cdot \frac{\sqrt{2}}{2} = 80$, so $y = 6 + 80t - 16t^2$.
- $y(2) = 6 + 160 - 64 = 102$ ft.
Answer: $102$ ft
Common mistake
Using the whole launch speed in both equations, such as $x = 64t$ and $y = 64t - 16t^2$. The ball does not move $64$ ft/s sideways and $64$ ft/s up; the speed is split between the two directions by $\cos\theta$ and $\sin\theta$. Check: $(v_0\cos\theta)^2 + (v_0\sin\theta)^2$ should equal $v_0^2$.