A Parachutist bails out from an aeroplaneand after dropping through a distance of 40m opens the parachute and decelerates at - Physics - Motion In A Straight Line

Question

A Parachutist bails out from an aeroplaneand after dropping
through a distance of 40m opens the parachute and decelerates at
2m/s2 if he reach the ground with a speed of 2m/s how long he is in
air? at what height did he bail out from the plane

Neha Gupta · Accepted Answer

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

A Parachutist bails out from an aeroplaneand after dropping through a distance of 40m opens the parachute and decelerates at 2m/s2 if he reach the ground with a speed of 2m/s how long he is in air? at what height did he bail out from the plane