A soldier jumps out from an aeroplane with a parachute .After dropping through a distance of 19.6 m he opens the parachute and decelarates at the rate of 1 m/s . If he reaches the ground with a speed of 4.6 m/s , how long was he in air/

After opening parachute.... a=-1m/s^2, v =4.6m/s , u1=?, t1=t... V - u1 = a*t1 ... 4.6 - u1 = -1*t1 ....u1 =( t1 + 4.6 ).... Before opening parachute ......u =0, v = u1 , a=9.8 , s = 19.6 .....v^2 - u^2 = 2as... v^2-0 = 2 *9.8*19.6 v = 19.6 m/S....u1 = 19.6 ....t1+ 4.6= 19.6....t1=15 sec...Now again...s = u*t2 + a*(t2)^2....s = 0 + 9.8*(t2)^2.....19.6=19.6*( t2)^2.....t2= 1 sec ...Total time taken = t1 +t2= 15 + 1 =16 sec.....

t1=time taken before opening parachuteu=0 ,h1=-19.6m distance travelled before opening parachuteNow from h=ut+1/2gt^2,we have-19.6=0-1/2×9.8t1^2=> t1^2=19.6/4.9=4Or t1=2 Let v1=velocity attained after falling through 19.6 mu=0Now from v^2-u^2=2ghv^2-0=2×9.8×19.6v1=19.6m/su=v1=19.6m/sv=4.6m/sa=-1m/s^2from v=u+at4.6=19.6-t2Or t2=19.6-4.6=15sec•°• t1+t2=2+15=17secLet h2=height through which soldier falls after opening the parachute.u=v1=19.6m/sv=4.6m/sa=-1m/s^2•°•from v^2-u^2=2ah(4.6)^2-(19.6)^2=-2×1×h2•°•h2=181.5mTotal height,h=h1+h2=19.6+181.5=201.1m.. Thank you