Projectile Motion

Time of Flight and Range

when and where it lands

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

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.

tland=2vyg=2 ⋅ 10 m/s10 m/s2=2 st_{\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.

vygtland5 m/s10 m/s21 s10 m/s10 m/s22 s15 m/s10 m/s23 s\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.

vygtland10 m/s5 m/s24 s10 m/s10 m/s22 s10 m/s20 m/s21 s\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=vx tland=5 m/s ⋅ 2=10 mx = v_x\,t_{\text{land}} = 5\ \text{m}/\text{s}\,\cdot\,2 = \hl{10}\ \text{m}
Where it landsThe ball's arc from the launch point back down to its landing spot on the ground, with the landing point marked.starttopland

Same flight time, more across speed, more range

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

vxtlandx5 m/s2 s10 m10 m/s2 s20 m15 m/s2 s30 m\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.

vxtlandx5 m/s1 s5 m5 m/s2 s10 m5 m/s3 s15 m\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}
Where it landsThe ball's arc from the launch point back down to its landing spot on the ground, with the landing point marked.starttopland
mechanics Symmetric up-and-down flight with clean numbers lands at a whole second, so the range is exact.