Projectile Motion

Decompose the Launch

across and up are separate

A projectile is two simple motions at once: a steady horizontal coast and a vertical throw under gravity. Splitting them is the whole trick.

mechanics Giving the launch as a clean horizontal part and vertical part lets each direction be solved on its own.

Example

A projectile is two simple motions at once: a steady horizontal coast and a vertical throw under gravity. Splitting them is the whole trick. Exact arithmetic here means exact results for the stated model inputs; measured inputs still carry uncertainty and significant-figure limits.

highlighted = computed this step

A ball thrown up and forward

A ball is launched with a horizontal part of 5 metres per second and a vertical part of 10 metres per second. The slanted launch arrow is the real velocity; we work in these two parts precisely so we never need its messy diagonal length. The trick is to treat across and up as two separate, simpler problems.

vx=5 m/s,vy=10 m/sv_x = 5\ \text{m}/\text{s}, \quad v_y = 10\ \text{m}/\text{s}
The launch velocity split into across and upThe launch arrow with its constant horizontal part and its upward vertical part drawn from the launch point.ballvxvylaunch

Across: nothing pushes it sideways

Gravity only pulls down, so nothing changes the across speed. The horizontal part stays 5 metres per second the whole flight: across position is just that speed times time, equal steps every second.

x=vx t=5 m/s ⋅ tx = v_x\,t = 5\ \text{m}/\text{s}\,\cdot\,t
Across alone: equal steps every secondThe sideways position by itself: equally spaced marks with equal arrows, because the across speed never changes.t=0t=1t=2

Same speed, more time, more across distance

Hold the across speed fixed and let time advance. The across distance grows by the same amount each second.

vxtx5 m/s0 s0 m5 m/s1 s5 m5 m/s2 s10 m\begin{array}{c|c|c}v_x & t & x \\ \hline 5\ \text{m}/\text{s} & 0\ \text{s} & 0\ \text{m} \\ 5\ \text{m}/\text{s} & 1\ \text{s} & 5\ \text{m} \\ 5\ \text{m}/\text{s} & 2\ \text{s} & 10\ \text{m}\end{array}

Same time, more across speed, more distance

Hold time fixed. A larger across speed gives a larger across distance in the same time.

vxtx5 m/s1 s5 m10 m/s1 s10 m15 m/s1 s15 m\begin{array}{c|c|c}v_x & t & x \\ \hline 5\ \text{m}/\text{s} & 1\ \text{s} & 5\ \text{m} \\ 10\ \text{m}/\text{s} & 1\ \text{s} & 10\ \text{m} \\ 15\ \text{m}/\text{s} & 1\ \text{s} & 15\ \text{m}\end{array}

Up: gravity slows it, stops it, drops it

The up part is free fall with a head start. Gravity takes 10 metres per second off the upward speed each second, so the climb slows to a stop at the top, then falls back. The vertical position is the launch-up minus the usual one-half g t squared.

y=vy t−12 g t2y = v_y\,t - \tfrac{1}{2}\,g\,t^{2}
Up alone: the climb slows toward the topThe upward position by itself on a vertical line: the gaps shrink as gravity slows the climb to a stop at the top.starttop

The vertical formula rises then returns

Keep the same launch-up speed and gravity. Reading the formula at successive seconds shows the height climb and then come back.

tvyy0 s10 m/s0 m1 s10 m/s5 m2 s10 m/s0 m\begin{array}{c|c|c}t & v_y & y \\ \hline 0\ \text{s} & 10\ \text{m}/\text{s} & 0\ \text{m} \\ 1\ \text{s} & 10\ \text{m}/\text{s} & 5\ \text{m} \\ 2\ \text{s} & 10\ \text{m}/\text{s} & 0\ \text{m}\end{array}

More upward launch speed gives more height

Hold the clock at the same instant. A larger launch-up speed leaves more height after gravity has pulled for that same time.

vyty5 m/s1 s0 m10 m/s1 s5 m15 m/s1 s10 m\begin{array}{c|c|c}v_y & t & y \\ \hline 5\ \text{m}/\text{s} & 1\ \text{s} & 0\ \text{m} \\ 10\ \text{m}/\text{s} & 1\ \text{s} & 5\ \text{m} \\ 15\ \text{m}/\text{s} & 1\ \text{s} & 10\ \text{m}\end{array}