text: # 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.​
  
  • 8
  
  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
  
  • 21
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