Without air, a thrown object follows a symmetric parabola. With air resistance, it slows down, falls more steeply than it rose, and lands much shorter. Try it below, then revise the ideas behind it.
Simulator
Object
- Range
- –
- Max height
- –
- Flight time
- –
- Landing speed
- –
- Terminal velocity
- –
Key ideas
- Two forces act in flight: weight (down, constant) and drag (opposite to the velocity, growing with speed).
- Drag grows with speed squared at everyday speeds: double the speed and the drag is four times bigger.
- Terminal velocity is reached when drag equals weight. The resultant force is zero, so the object stops accelerating.
- Horizontal speed only ever falls with air resistance, because drag always has a backward component. That is why the path is lopsided and the fall is steep.
- Light, draggy objects (a shuttlecock has a terminal velocity of about 6.8 m/s) lose speed fast: a smash leaving at 80 m/s is under 20 m/s by the time it lands.
- Without air resistance, horizontal and vertical motion are independent, the path is a parabola, and the best angle for range on flat ground is 45°.
Equations
Weight
W = mgNewton’s second law
F = maDrag force
Fd = ½ρCdAv²Terminal velocity
vt = √(2mg ÷ ρCdA)No air: range on flat ground
R = u²sin2θ ÷ gNo air: time of flight
t = 2u sinθ ÷ gSUVAT
v = u + at · s = ut + ½at² · v² = u² + 2asWorked example
A shuttlecock has mass 5.0 g and a terminal velocity of 6.8 m/s. At terminal velocity, drag = weight, so Fd = mg = 0.005 × 9.81 = 0.049 N. Because drag ∝ v², at 13.6 m/s (twice vt) the drag is 4 × 0.049 = 0.196 N. Falling straight down at that speed, the resultant force is 0.196 − 0.049 = 0.147 N upward, so it decelerates at 0.147 ÷ 0.005 = 29 m/s² (about 3g).