A) 2 s
B) 4 s
C) 5 s
D) 7 s
Correct Answer: C
Solution :
As ball falls from rest, hence, \[\operatorname{u} =0\] Now, distance covered in the last second is equal to distance covered in first 3 s of motion. \[\therefore \,\,\,\,\,\,\,\,\,\,\,\,\,\,{{S}_{3}}={{S}_{n}}\] [\[{{S}_{3}}\]= distance covered in first 3 s \[{{S}_{n}}\] = distance covered in last second] \[{{S}_{3}}=\,\,0+\frac{1}{2}g{{t}^{2}}=\,\,\frac{1}{2}\times 10\times 9=45\] Now, \[{{S}_{n}}=45\] \[\Rightarrow \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,{{S}_{n}}=\,\,u+\frac{g}{2}(2t-1)\] \[45=0+\frac{10}{2}\left( 2t-1 \right)\] \[45=5\left( 2t-1 \right)\] \[9+1=2t\] \[t=5\] \[\Rightarrow \,\,\,\,\,\,\,\,\,\,\,\,\,t=5s\]
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