Question

Questions 1. A parachutists bails out from an aeroplane and after dropping through a distance of 40m opens the parachute and decelerates at 2 m/s sq. If he reaches the ground with a speed of 2m/s, how long is he in the air? At what height did he bail out from the plane.

Answer 1

Time taken to cover 40m after bailing out, $t_{1}=\sqrt{\frac{2h}{g}}=4\sqrt{\frac{5}{g}}$.
Velocity at that time, $u=-gt_{1}=-g\sqrt{\frac{80}{g}}=-4\sqrt{5g}$.
Time taken to cover remaining distance, $t_{2}=\frac{v-u}{a}=\frac{(-2)-(-4\sqrt{5g})}{+2}=-1+2\sqrt{5g}$.
Total time, $T=t_{1}+t_{2}=4\sqrt{\frac{5}{g}}+2\sqrt{5g}-1$.
Put the value of 'g' you get your answer.

Answer 2

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