The parachutist jumped from a height of 2 km. Before opening the parachute he flew at a speed of 50 m / s

| August 12, 2021 | education

The parachutist jumped from a height of 2 km. Before opening the parachute he flew at a speed of 50 m / s, after opening – at a speed of 5 m / s, and the average speed of his movement was equal to 6.45 m / s. How many seconds after the start of the jump he opened the parachute ?

h = 2 km = 2000 m.

V1 = 50 m / s.

V2 = 5 m / s.

Vav = 6.45 m / s.

t1 -?

Let us write down the definitions for the average flight speed Vav: Vav = h / t, where t is the time of the entire flight.

t = h / Vav.

t = 2000 m / 6.45 m / s = 310 s.

t = t1 + t2, where t1 is the flight time before the parachute opens, t2 is the flight time after the parachute opens.

t1 = h1 / V1.

t2 = h2 / V2 = (h – h1) / V2.

h1 / V1 + (h – h1) / V2 = t.

(h1 * V2 + (h – h1) * V1) / V1 * V2 = t.

h1 * V2 + h * V1 – h1 * V1 = V1 * V2 * t.

h1 * (V2 – V1) = V1 * V2 * t – h * V1.

h1 = (V1 * V2 * t – h * V1) / (V2 – V1).

h1 = (50 m / s * 5 m / s * 310 s – 2000 m * 50 m / s) / (5 m / s – 50 m / s) = 500 m.

t1 = 500 m / 50 m / s = 10 s.

Answer: the flight time before the parachute is deployed is t1 = 10 s.



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