Question

a parachutist bails out from an aeroplane and after dropping throw distance of 40 m he opens the parachute and D accelerates at 2 metre per second square if he reaches the ground with a speed of 2m per second how long is he in the air at what height did he will out from the plane

Answer 1

Initial velocity u = 0
Travels 40 m in time t with acceleration g.

s = ut +1/2at^2=>40 = 0+1\2×10×t^2t = 2.82 s

velocity after time t

v = u + gt

v = 0 + 10×2.82

v = 28.2 m/s

When prachutist oens the parachute, intial velocity is 28.2 m/s
deaccelerates at 2 m/s2 and reach to the ground with 2 m/s. hence final velocity is 2m/s2 = 28.2 − 2t1

where t1 is the time taken to reach the gound.t1 = 13.1 s

distance travel in t1 ss1 = ut1 + 1/2at12 = 28.2×13.1 − 12×2×13.1×13.1= 197.81m

total time taken = 13.1+2.82 = 15.92 s

height from ground = 197.81 +40 = 237.81 m

@skb