A) 8 m/s
B) 6 m/s
C) 10 m/s
D) data is not sufficient
Correct Answer: C
Solution :
Key Idea s Velocity is the rate of change of displacement. Velocity of particle, \[v=\]rate of change of displacement i.e., \[v=\frac{dr}{dt}\] Given, \[x=6t\] ?(i) and \[y=8t-5{{t}^{2}}\] ?(ii) Differentiating Eqs.(i) and (ii), we get \[{{v}_{x}}=\frac{dx}{dt}=\frac{d}{dt}(6t)=6\,m/s\] and \[{{v}_{y}}=\frac{dy}{dt}=\frac{d}{dt}(8t-5{{t}^{2}})=(8-10)m/s\] At \[t=0,\] \[{{v}_{x}}{{|}_{t=0}}=6\,m/s\] and \[{{v}_{y}}{{|}_{t=0}}=8\,m/s\] Hence, velocity of projection at time \[t=0\]is \[v=\sqrt{{{v}_{x}}^{2}+{{v}_{y}}^{2}}=\sqrt{{{(6)}^{2}}+{{(8)}^{2}}}\] \[=\sqrt{36+64}=\sqrt{100}=10\,m/s\]
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