question_answer

A parachutist drops freely from an aero plane for 10 s and then the parachute opens out. Then he descends with a net retardation of\[\text{12 m}{{\text{s}}^{-\text{2}}}\]. If he strikes the ground with a velocity of\[~\text{2}0\text{ m}{{\text{s}}^{-\text{1}}},\]what is the height at which he gets out of the plane?

Answer:

Velocity acquired after 10 seconds is \[u=gt=98\,m{{s}^{-1}}\] Velocity on reaching the ground is\[v=20\,\,m{{s}^{-1}}\]retardation\[=12\,\,m{{s}^{-2}}\]distance travelled after parachute opened \[={{s}_{2}}\]from\[{{v}^{2}}-{{u}^{2}}=2a{{s}_{2}}\] \[{{20}^{2}}-{{98}^{2}}=2(-12){{s}_{2}}\] \[118\times 78=2\times 12\times {{s}_{2}}=383.5\,m\] Before parachute open, the distance travelled by him is\[{{s}_{1}}=\frac{1}{2}g{{t}^{2}}\] i.e., \[{{s}_{1}}=\frac{1}{2}\times 9.8\times 100=490\,m\] It means the height at which he gets out of the plane is \[{{s}_{1}}+{{s}_{2}}=873.5\,m\]