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Q.A parachutist bails out from an aero plane and after dropping through a distance of 40m, he opens his parachute and decelerates at 2m/s. If he reaches the ground with a speed of 2m/s, how long is in air? At what height did he bail out of the plane?

Initial velocity u = 0
Travels 40 m in time t with acceleration g.​s = ut +12at240 = 0+12×10×t2t = 2.82 svelocity after time tv = u + gtv = 0 + 10×2.82v = 28.2 m/sWhen prachutist oens the parachute, intial velocity is 28.2 m/sdeaccelerates at 2 m/s2  and reach to the ground with 2 m/s. hence final velocity is 2m/s2 = 28.2 - 2t1where t1 is the time taken to reach the gound.t1 = 13.1 sdistance travel in t1 ss1 = ut1 + 12at12 = 28.2×13.1 - 12×2×13.1×13.1= 197.81mtotal time taken = 13.1+2.82 = 15.92 sheight from ground = 197.81 +40 = 237.81 m

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u = 0, g = 9.8 ms-2, s= 40m , t=?, v=?

s= ut + 1/2 gt2

40= 0 + 1/2 * 9.8* t2

t= root 80/9.8 = 20/7s = 2.86s

v = u+gt = 0+9.8*20/7 = 28ms-1

when the parachutist decdelarates uniformly

u= 28 m/s , a= -2 ms-2 ,v= 2 m/s

t= v-u/t = 2-28/-2 = 13s

s= ut+1/2at2 =28*13- 1/2*2*132

= 364-169 = 195 m

total time in air = 2.86+13 = 15.86 s

height at which parachutist bails out = 40+195 = 235m

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