A) \[\text{142 m }{{\text{s}}^{\text{-2}}}\]
B) \[\text{285 m }{{\text{s}}^{\text{-2}}}\]
C) \[\text{338 m }{{\text{s}}^{\text{-2}}}\]
D) \[\text{564 m }{{\text{s}}^{\text{-2}}}\]
Correct Answer: C
Solution :
Here height from which ball is dropped h = 20 m Using\[{{\text{v}}^{\text{2}}}\text{=}{{\text{u}}^{\text{2}}}\text{+2gh;}\]\[{{\text{v}}^{2}}=2gh(\because u=0)\] or \[v=\sqrt{2gh}\] (downward) Ball rebounds to a height, h' = 10 m So, \[{{0}^{2}}={{v}^{'2}}-2gh'\] or \[v'=\sqrt{2gh'}\] (upward) Acceleration of the ball, \[a=\frac{(v'-v)}{t}=\frac{[\sqrt{2gh'}-(-\sqrt{2gh})]}{t}\] Or \[a=\frac{[\sqrt{2\times 9.8\times 10}+\sqrt{2\times 9.8\times 20}]}{0.1}=338\text{ m }{{\text{s}}^{-2}}\]
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