A) \[\sqrt{8}\,\,s\]
B) \[2\,\,s\]
C) \[4\,\,s\]
D) \[8\,\,s\]
Correct Answer: C
Solution :
We know\[h=ut+\frac{1}{2}g{{t}^{2}}\] Since, vertical velocity,\[u=0\] and \[h=7848\,\,cm\] \[\therefore \] \[7898=0+\frac{1}{2}\times 981\times {{t}^{2}}\] \[\Rightarrow \] \[{{t}^{2}}=\frac{7848\times 2}{981}=8\times 2=16\] \[\Rightarrow \] \[t=4\,\,s\] Note: The time taken by the particle to reach the ground is \[t=\sqrt{\frac{2h}{g}}\]You need to login to perform this action.
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