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NAGŁÓWEK

Instructions — NAGŁÓWEK

Odpowiedź:

Kąt 45° jest optymalny tylko gdy wysyłasz z poziomu gruntu i lądujesz na tym samym poziomie bez oporu powietrza.

Gdy wysyłasz z wysokości h₀ > 0 (np. rzut koszykarski), optymalny kąt maleje:

$$\theta_{\text{opt}} = \arctan\left(\frac{v_0}{\sqrt{v_0^2 + 2gh_0}}\right) < 45°$$

Przykład: Dla h₀ = 2 m, v₀ = 10 m/s, g = 9.81 m/s²:

$$\theta_{\text{opt}} \approx 40.2°$$

Formulas

Odpowiedź:

Opór powietrza jest znacznie ważniejszy, niż myśli większość ludzi:

  • Piłka do golfa (małą, gładką): Redukcja ~30–40% zasięgu
  • Piłka tenisowa (większa, puchata): Redukcja ~35–50%
  • Baseball (gładkie, ciężkie): Redukcja ~15–25%
  • Poperć futbolówkę (długa, obrót): Redukcja + efekty boczne

Empirycznie, dla średnich prędkości sportowych, opór powietrza zmniejsza zasięg o 20–40%.

Article — NAGŁÓWEK

Projectile Motion Calculator

Projectile motion describes a body launched into the air and acted on only by gravity. Horizontal motion is uniform at v₀ cos θ; vertical motion accelerates downward at g. The trajectory is a parabola described by y = h₀ + x tan θ − g x² / (2 v₀² cos² θ).

Galileo published this analysis in 1638 in Two New Sciences. He had figured it out decades earlier by rolling balls down inclined planes and timing them with a water clock. The work overturned the Aristotelian idea that projectiles moved on two separate paths — first a straight push, then a vertical fall — and established the parabolic arc as the correct geometry of ballistic flight.

What is projectile motion?

Projectile motion is the curved path of any object launched with an initial velocity, once it is in flight and subject only to gravity. The classic idealization ignores air drag, wind, the Coriolis force, and the curvature of the Earth. With those simplifications, every projectile from a thrown ball to a cannonball follows the same parabolic shape, scaled by initial velocity, angle, and gravity.

The two independent components — horizontal and vertical — make the problem tractable. Horizontal velocity stays constant. Vertical velocity decreases at g during the rise, reaches zero at the apex, then increases downward during the fall. By solving the time equations separately, you can predict where the projectile lands, when, and at what speed.

Did you know

The longest verified golf drive is 515 yards (471 m), by Mike Austin in 1974 at the U.S. Senior Open in Las Vegas. The strong tailwinds and dry desert air helped, but the carry was still well below the vacuum-parabola prediction for an exit velocity of ~80 m/s — direct evidence that air drag costs roughly half the theoretical range.

The projectile motion formulas

From the position equations x(t) = v₀ cos θ · t and y(t) = h₀ + v₀ sin θ · t − g t² / 2, three derived quantities matter most. Range R is the horizontal distance to impact. Maximum height H is the peak of the trajectory. Time of flight is how long the projectile is airborne.

For a level launch (h₀ = 0), R = v₀² sin(2θ) / g, H = (v₀ sin θ)² / (2g), and t = 2 v₀ sin θ / g. A 25 m/s, 45° throw on Earth gives R = 63.7 m, H = 15.9 m, t = 3.6 s. The same numbers on the Moon become R = 386 m, H = 96.5 m, t = 21.8 s — six times the range and time, because Moon gravity is one-sixth of Earth's.

Vacuum projectile shortcuts
R = v₀² sin 2θ / g h₀ = 0
H = v₀² sin²θ / 2g peak height
t = 2 v₀ sin θ / g flight time
θ_opt = 45° level ground

Projectile motion range and the 45° myth

The "45° gives maximum range" rule is half right. It holds only when launch and landing are at the same height and there is no air resistance. Once you launch from a height above the target — basketball, shot put, javelin — the optimum drops. For a 2 m release point and 10 m/s throw, the optimum is about 40°. For a shot put, where the release is 2.2 m above the toe board, the optimum is ~41–43°.

Even on level ground with no drag, the broader range of "good" angles is wider than people assume. A 30° launch and a 60° launch produce the same range (different times of flight). Athletes who train for distance often pick the lower angle because the projectile spends less time in the air and is less affected by wind.

Vacuum results overestimate real range

This calculator ignores air drag. For a baseball thrown at 40 m/s, the vacuum range at 45° is 163 m, but real-world maxima are around 130 m — a 20% reduction. For a golf ball at 75 m/s, vacuum predicts 574 m; real drives top out near 350 m. Use vacuum numbers as the upper bound, not the answer.

Projectile motion on other planets

Surface gravity directly scales every projectile result. Moon gravity is 1.62 m/s², Mars 3.71 m/s², Venus 8.87 m/s², Jupiter (at the cloud tops) 24.79 m/s². On the Moon, the Apollo 14 astronaut Alan Shepard hit two golf balls in 1971 with a makeshift 6-iron. He reported the second went "miles and miles" — modern image analysis of the 16 mm film puts the actual distance at about 37 yards (33.8 m), unspectacular until you correct for the bulky pressure suit limiting his swing.

Mars is the next likely target for human projectile experiments. At g = 3.71, the same launch flies 2.64× as far. Future Martian recreation could include hammer throwing, javelin, and long-distance baseball — all with much-extended ranges that would force new field dimensions and rule sets.

Projectile motion with air resistance

Real projectiles bleed energy to drag. The drag force is F = ½ ρ v² C_d A, where ρ is air density (~1.225 kg/m³ at sea level), C_d is the drag coefficient (typically 0.4–1.0 for sports balls), and A is the cross-sectional area. Drag grows with v², so high-speed projectiles lose proportionally more to it than slow ones.

Drag also shifts the optimum launch angle below 45°. For a tennis ball at 60 m/s, the drag-corrected optimum is about 28°. For artillery at 600 m/s, drag is so dominant that ballistic tables are computed numerically — the parabolic approximation is unusable beyond a few hundred meters.

  • Air density 1.225 kg/m³ at sea level, drops to about 0.4 kg/m³ at the cruise altitude of an airliner
  • Drag coefficient 0.47 for a smooth sphere, ~0.3 for a dimpled golf ball
  • Magnus force on a spinning ball can curve trajectory by ~5 m over a 30 m throw
  • Wind changes effective launch velocity by up to ±15% in open stadiums
  • Modern ballistic tables use numerical integration with table-lookup drag

Projectile motion in sports

Different sports converge on different launch angles. Long jump optimum is about 21° because the runner cannot generate more vertical speed without losing horizontal speed. Shot put optimum is 41° because the heavy iron ball has low drag and benefits from the longer time aloft. Golf optimum is 12–18° because backspin generates lift and a higher launch reduces it.

Baseball pitching uses different physics again. A 95 mph fastball drops 1.0 m between mound and plate; without the late curveball spin from a 12-6 break, it would drop more. Catcher framing, hitter timing, and umpire judgment all depend on milliseconds derived from these same projectile equations.

Tip

To estimate a projectile's range in your head, square the velocity in m/s and divide by 10. A 20 m/s launch at 45° on Earth gives roughly 400 / 10 = 40 m (true: 40.8 m). The error is under 2% for the level-ground case.

Common projectile motion mistakes

The first is forgetting to convert velocity. Speeds in km/h must be divided by 3.6 before entering them as m/s. The second is using degrees where radians are expected — most calculators and spreadsheets default to radians for sin and cos. The third is treating the launch height as the impact height when they differ; this changes the range formula significantly.

The fourth is assuming the 45° rule applies everywhere. It does not. Once h₀ ≠ 0, the optimum drops. Once air drag matters, it drops further. The fifth is reading impact velocity wrong: the impact speed for level launch equals the launch speed, but impact angle is the negative of launch angle, not the same.

Odpowiedź:

Maksymalna wysokość jest osiągana, gdy składowa pionowa prędkości = 0:

$$H_{\max} = h_0 + \frac{(v_0 \sin(\theta))^2}{2g}$$

Przykład: h₀ = 0, v₀ = 20 m/s, θ = 45°:

$$H_{\max} = 0 + \frac{(20 \times 0.707)^2}{2 \times 9.81} = \frac{(14.14)^2}{19.62} \approx 10.2 \text{ m}$$

Dla θ = 90° (wysłanie pionowe):

$$H_{\max} = \frac{v_0^2}{2g} = \frac{400}{19.62} \approx 20.4 \text{ m}$$