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How does the speed parachute help you?

Displacement, velocity and acceleration (please help!)?

  • Could you please explain how to solve this question? A parachutist drops from an aeroplane so that the constant acceleration during free fall due to gravity and air resistance is 8 m/s2. The parachute is released after 6 seconds, uniformly retarding the parachutist in 28 seconds to a constant speed of 2.5 m/s. This speed is maintained until the parachutist reaches the ground which is 1101 metres below the point of release. a How long is the parachutist in the air? b After how long has the parachutist fallen half the distance (answer to the nearest tenth of a second)? Answers : a)2 min 14 s b)16.2 s Thanks for any help!

  • Answer:

    a) Acceleration phase a = 8 m/sec^2 t1 = 6.0 sec V1 = a*t1 = 8*6 = 48 m/sec h1 = 1/2V1*t1 = 24*6 = 144 m b) Deceleration phase V1 = 48 m/sec V2 = 2,5 m/sec t2 = 28 sec h2 = (V1+V2)/2*t2 = 50.5*14 = 707 m dec = (V1-V2)/t2 = 45.5/28 = 1.625 m/sec^2 c) Constant speed phase h = 1101 m h3 = h-(h1+h2) = 1101-851 = 250m t3 = h3/V2 = 250/2.50 = 100.0 sec total time t in the air : t = t1+t2+t3 = 100+28+6 = 134 sec time t4 to stretch half distance from drop.... h/2 = 1101/2 = 550,5 m h/2-h1 = 550,5-144 = 406.50 406.50 = (V1+(V1-dec*t4)*t4/2 813 = (48+48-1.625t4)*t4 813-96t4+1.625t4^2 = 0 t4 = (96±√96^2+813*4*1.625)/3.25 = (96+120.4)/3.25 = 66.59 sec

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distance in first 6s: y1 = (1/2)(a1)(t1)² y1 = (1/2)(8 m/s²)(6 s)² y1 = 144 m velocity when chute opens: v1 = (a1)(t1) v1 = (8 m/s²)(6 s) v1 = 48 m/s acceleration in next 28 s: v2 = v1 + (a2)(t2) 2.5 m/s = 48 m/s + (a2)(28 s) a2 = -1.625 m/s² distance in next 28s: y2 = (v1)(t2) + (1/2)(a2)(t2)² y2 = (48 m/s)(28 s) + (1/2)(-1.625 m/s²)(28 s)² y2 = 707 m additional distance to reach the ground: y3 = 1101 m - y1 - y2 y3 = 1101 - 144 - 707 m y3 = 250 m additional time to reach the ground: y3 = (v2)(t3) 250 m = (2.5 m/s)(t3) t3 = 100 s total time: t = t1 + t2 + t3 t = 6 + 28 + 100s t = 134s (2 min 14 sec) According to the distances already calculated, the half-distance mark would occur while the chute is opening. y4 = y1 + (v1)(t4) + (1/2)(a2)(t4)² 1101 m / 2 = 144 m + (48 m/s)(t4) + (1/2)(-1.625 m/s²)(t4)² 0.8125(t4)² - 48(t4) + 406.5 = 0 t4 = [48 - sqrt[(-48)² - 4(0.8125)(406.5)]] / [2(0.8125)] t4 = 10.2 s t1 + t4 = 6 + 10.2s = 16.2s

DR

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