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# A parachutist bails out from an aeroplane and after drooping through a distance of 40 m, he opens the parachute and decelerates at 2 m/s2. If he reaches the ground with a speed of 2 m/s (i) how long he is in air. (ii) at what height did he bail out from the plane?

Solution
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#### Applying 3rd equation to calculate the final velocity, v2=0+2∗9.8∗40v = √784 = 28 m/s . Time taken, t1=289.8 [ u + at ] = 2.85 s.Then, the second part says that he reached the ground with velocity 2 m/s (final velocity) with deceleration of 2 m/s2 and the initial velocity would be which we calculated above i.e. 28m/sAssume, it traveled a distance of 'h' while falling.Again, apply 3rd equation of motion, (2)2=(28)2+2(−2)∗h4 * v = 784 - 4 = 780 h = 195m.Time taken, t2=28−22 [ v = u + at ] = 13 s Time taken while he is in air, t = t1+t2 = 2.85 + 13 = 15.85 s

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