question_answer

A parachutist after bailing out falls 50 m without friction. When parachute opens, it decelerates at 2 m/s2. He reaches the ground with a speed of 3 m/s. At what height, did he bail out?                                    [AIEEE 2005]           

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.