question_answer

The parachute got opened 12 seconds after a person had jumped out of an aeroplane. If the parachute descends with a retardation of\[\text{2 m}/{{\text{s}}^{\text{2}}}\]and if he falls out of the plane when it is at a height of 4220 m, what is the velocity with which he reaches the ground?                     [\[\text{g}=\text{1}0\text{ m}/{{\text{s}}^{\text{2}}}\]]

Answer:

He falls freely for 12 seconds and the distance covered by him is: \[S=\frac{1}{2}g{{t}^{2}}=\frac{1}{2}\times 10\times 144=720\,m\] He falls 4220 ? 720 = 3500 m with retardation of \[2\,m/{{s}^{2}}\] Velocity when retardation starts = gt \[=10\times 12=120\,\,m/s\] This becomes the initial velocity per the second part. Let v be the final velocity with which he reaches the ground. \[\Rightarrow {{v}^{2}}-{{u}^{2}}=2aS\] \[{{v}^{2}}-{{(120)}^{2}}=2(-2)3500\] \[{{v}^{2}}={{(120)}^{2}}-14000=400\,\,or\,\,v=20\,m/s\]