A) 2m
B) 6m
C) 8m
D) 10m
Correct Answer: D
Solution :
Upward acceleration\[=\frac{5}{5}=1\text{ }m/{{s}^{2}}\] Upward distance covered in 4 s \[y=\frac{1}{2}a{{t}^{2}}\] \[=\frac{1}{2}\times 1\times {{(4)}^{2}}=8\,m\] Horizontal distance covered in 4 s \[x=vt=1.5\times 4=6m\] \[\therefore \] \[s=\sqrt{{{x}^{2}}+{{y}^{2}}}=\sqrt{{{6}^{2}}+{{8}^{2}}}\] \[=\sqrt{36+64}=10\,m\]
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