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Airplane dropping water on fire

Short answer

Convert speed: 315 km/h = 87.5 m/s. Time to fall 80 m: t = √(2×80/9.81) = 4.04 s. Release distance d = 87.5 × 4.04 ≈ 353 m.

An airplane used to drop water on brushfires is flying horizontally in a straight line at 315 km/h at an altitude of 80 m. Determine the distance dd at which the pilot should release the water so that it will hit the fire at B.

Once released, the water moves as a projectile with the plane's horizontal velocity and zero initial vertical velocity.

Data

vx=315 km/h=3153.6=87.5 m/s,h=80 m,g=9.81 m/s2v_x=315\ \text{km/h}=\frac{315}{3.6}=87.5\ \text{m/s},\qquad h=80\ \text{m},\qquad g=9.81\ \text{m/s}^2

Time to fall 80 m

h=12gt2 ⇒ t=2(80)9.81=4.0386 sh=\tfrac12 g t^{2}\ \Rightarrow\ t=\sqrt{\frac{2(80)}{9.81}}=4.0386\ \text{s}

Horizontal distance

d=vxt=87.5(4.0386)=353.4 md=v_x t=87.5(4.0386)=353.4\ \text{m}

Answer: the water must be released d≈353d\approx 353 m before the fire (horizontal distance).

✓ Answer checked by Claude Opus 5.5 (high effort)

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