question_answer

The height of a tower is\[7848\,\,cm\]. A particle is thrown from the top of the tower with the horizontal velocity of \[1784\,\,cm/s\]. The time taken by the particle to reach the ground is (g \[=981\,\,cm/{{s}^{2}})\]:

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                 \[t=\sqrt{\frac{2h}{g}}\]