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 \[{{s}_{1}}+{{s}_{2}}=873.5\,m\]
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