A) 10 m
B) 8 m
C) 6 m
D) 2 m
Correct Answer: A
Solution :
\[{{S}_{\text{Horizontal}}}=ut=1.5\times 4=6\,m\] \[{{S}_{\text{Vertical}}}=\frac{1}{2}a{{t}^{2}}=\frac{1}{2}\frac{F}{m}{{t}^{2}}=\frac{1}{2}\times 1\times 16=8\,m\] \[{{S}_{\text{Net}}}=\sqrt{{{6}^{2}}+{{8}^{2}}}=10\,m\]
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