Question

a parachute bailing from an aeroplane after dropping through a distance of 40m opens the parachute and deaccelarates at 2m/sec2 if it reaches the ground with a speed of 2m/sec how long he is in the air & at what height did he bail out from plane

Answer 1

Draw the diagram. Let the distance of fall after 40m to ground be h. g=9.8 m/s2 FOR FIRST 40 metres u=0;h=40; v=(2gh)^[1/2] v=28 t1=v-u/a t1=28/9.8. t1=2.9. FROM 40m TO GROUND u=28;v=2;a=-2 t2=(v-u)/a t2=13 sec s=ut+1/2at^2 s=28*13-1/2*(-2)13*13 s=13*15 s=195m Therefore total time =2.9+13=15.9 sec Total height =40+195m==235m