From a stationary helicopter located at an altitude of 1000m, a pistol was fired vertically downward, and the bullet flies

| September 18, 2021 | education

From a stationary helicopter located at an altitude of 1000m, a pistol was fired vertically downward, and the bullet flies out at a speed of 200 m / s. In how many seconds and at what speed will the bullet reach the Earth’s surface?

Data: h (altitude at which the stationary helicopter is located; bullet path) = 1000 m; V0 (bullet velocity after firing) = 200 m / s.

Constants: g (acceleration due to gravity) ≈ 10 m / s2.

1) Bullet velocity at the surface: S = h = (V ^ 2 – V0 ^ 2) / 2a = (V ^ 2 – V0 ^ 2) / 2g, whence V = √ (h * 2g + V0 ^ 2) = √ (1000 * 2 * 10 + 200 ^ 2) = 244.95 m / s.

2) Duration of bullet movement: t = (V – V0) / a = (V – V0) / g = (244.95 – 200) / 10 = 4.5 s.

Check: S = h = (V + V0) * t / 2 = (244.95 + 200) * 4.5 / 2 ≈ 1000 m (correct).



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