A) 293 m
B) 111 m
C) 91 m
D) 182 m
Correct Answer: A
Solution :
After bailing out from point A parachutist falls freely under gravity. The velocity acquired by it will ?v? From \[{{v}^{2}}={{u}^{2}}+2as\]\[=0+2\times 9.8\times 50\]= 980 [As u = 0, \[a=9.8m/{{s}^{2}}\], s = 50 m] At point B, parachute opens and it moves with retardation of 2\[m/{{s}^{2}}\] and reach at ground (Point C) with velocity of \[3m/s\] For the part ?BC? by applying the equation \[{{v}^{2}}={{u}^{2}}+2as\] \[v=3m/s\], \[u=\sqrt{980}m/s\], \[a=-2m/{{s}^{2}}\], s = h Þ \[{{(3)}^{2}}={{(\sqrt{980})}^{2}}+2\times (-2)\,\times \,h\]Þ \[9=980-4h\] Þ \[h=\frac{980-9}{4}\] \[=\frac{971}{4}=242.7\tilde{=}243\]m. So, the total height by which parachutist bail out = \[50+243\] = 293 m.You need to login to perform this action.
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