A) 3 s
B) 5 s
C) 7 s
D) 9 s
Correct Answer: B
Solution :
[b] \[{{S}_{nth}}=u+\frac{a}{2}(2n-1)\] \[=0+\frac{g}{2}(2n-1)\] \[=\frac{10}{2}(2n-1)\] \[=5(2n-1)\] Also. \[{{S}_{3}}=0+\frac{1}{2}g\times {{(3)}^{2}}\] \[=\frac{10}{2}\times 9=45\] As per question, \[{{S}_{nth}}={{S}_{3}}\] \[\therefore 5\text{ }\left( 2n-3 \right)=45\] or \[2n-1=9\] or n = 5 So, the body has fallen for a time of 5s.You need to login to perform this action.
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