A) \[t{{ & }_{1}}>{{t}_{2}}\]
B) \[t{{ & }_{1}}={{t}_{2}}\]
C) \[t{{ & }_{1}}<{{t}_{2}}\]
D) Information is insufficient
Correct Answer: A
Solution :
When the elevator is moving uniformly \[{{\text{v}}_{\text{coin,}\,\text{elevator}}}\text{=a}{{\,}_{\text{coin,}}}_{\text{ground}}\text{=g(downward)}\] \[{{\text{v}}_{\text{coin,}\,\text{elevator}}}\text{=0}\] \[{{s}_{\text{coin,}\,\text{elevator}}}\text{=h(downward)}\] from \[s=ut+\frac{1}{2}a{{t}^{2}}\]with elevator For \[h=0+\frac{1}{2}g_{1}^{2}\Rightarrow {{t}_{1}}=\sqrt{\frac{2h}{g}}\] When elevator is accelerating up with an acceleration a \[{{a}_{\text{coin,}\,\text{elevator}}}\text{=a+g(downward)}\] \[{{u}_{\text{coin,}\,\text{elevator}}}\text{=0}\] \[{{s}_{\text{coin,}\,\text{elevator}}}\text{=h(downward)}\] so,\[h=0+\frac{1}{2}(a+g)t_{2}^{2}\Rightarrow {{t}_{2}}=\sqrt{\frac{2h}{a+g}}\] so,\[{{t}_{1}}>{{t}_{2}}\]You need to login to perform this action.
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