Time of Flight and Range - Mechanics Step by Step

Because the across speed is constant, once we know how long the ball is in the air the distance falls right out.

mechanics Symmetric up-and-down flight with clean numbers lands at a whole second, so the range is exact.

Example

Because the across speed is constant, once we know how long the ball is in the air the distance it travels falls right out. Exact arithmetic here means exact results for the stated model inputs; measured inputs still carry uncertainty and significant-figure limits.

highlighted = computed this step

When does it land?

It lands when the height is back to zero. The up part climbs and falls symmetrically, so it lands after twice the time to the top. Gravity erases the 10 metres per second of upward speed at 10 per second, reaching the top at one second and landing at 2 seconds.

t_{\text{land}} = \frac{2 v_y}{g} = \frac{2\,\cdot\,10\ \text{m}/\text{s}}{10\ \text{m}/\text{s}^{2}} = \hl{2}\ \text{s}

More upward launch speed keeps it airborne longer

Hold gravity fixed. More upward launch speed takes longer to erase, so the whole up-and-down flight lasts longer.

\begin{array}{c|c|c}v_y & g & t_{\text{land}} \\ \hline 5\ \text{m}/\text{s} & 10\ \text{m}/\text{s}^{2} & 1\ \text{s} \\ 10\ \text{m}/\text{s} & 10\ \text{m}/\text{s}^{2} & 2\ \text{s} \\ 15\ \text{m}/\text{s} & 10\ \text{m}/\text{s}^{2} & 3\ \text{s}\end{array}

Stronger gravity shortens the flight

Hold the launch-up speed fixed. Larger gravity erases that upward speed faster, so landing comes sooner.

\begin{array}{c|c|c}v_y & g & t_{\text{land}} \\ \hline 10\ \text{m}/\text{s} & 5\ \text{m}/\text{s}^{2} & 4\ \text{s} \\ 10\ \text{m}/\text{s} & 10\ \text{m}/\text{s}^{2} & 2\ \text{s} \\ 10\ \text{m}/\text{s} & 20\ \text{m}/\text{s}^{2} & 1\ \text{s}\end{array}

How far does it go?

The across speed never changed, so the range is just 5 metres per second times the 2 seconds of flight: 10 metres.

x = v_x\,t_{\text{land}} = 5\ \text{m}/\text{s}\,\cdot\,2 = \hl{10}\ \text{m}

Same flight time, more across speed, more range

Hold the flight time fixed. Faster sideways coasting lands farther away.

\begin{array}{c|c|c}v_x & t_{\text{land}} & x \\ \hline 5\ \text{m}/\text{s} & 2\ \text{s} & 10\ \text{m} \\ 10\ \text{m}/\text{s} & 2\ \text{s} & 20\ \text{m} \\ 15\ \text{m}/\text{s} & 2\ \text{s} & 30\ \text{m}\end{array}

Same across speed, more time, more range

Hold the across speed fixed. More time in the air gives more across distance.

\begin{array}{c|c|c}v_x & t_{\text{land}} & x \\ \hline 5\ \text{m}/\text{s} & 1\ \text{s} & 5\ \text{m} \\ 5\ \text{m}/\text{s} & 2\ \text{s} & 10\ \text{m} \\ 5\ \text{m}/\text{s} & 3\ \text{s} & 15\ \text{m}\end{array}