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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+29.840

v = 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/s

Assume, it traveled a distance of 'h' while falling.

Again, apply 3rd equation of motion,

(2)2=(28)2+2(2)h

4 * v = 784 - 4

= 780

h = 195m.

Time taken, t2=2822 [ 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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