A) 8 m/sec
B) 6 m/sec
C) 10 m/sec
D) Not obtainable from the data
Correct Answer: C
Solution :
\[{{v}_{y}}=\frac{dy}{dt}=8-10t\], \[{{v}_{x}}=\frac{dx}{dt}=6\] at the time of projection i.e. \[{{v}_{y}}=\frac{dy}{dt}=8\]and \[{{v}_{x}}=6\] \[\therefore v=\sqrt{v_{x}^{2}+v_{y}^{2}}=\sqrt{{{6}^{2}}+{{8}^{2}}}=10\ m/s\]You need to login to perform this action.
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