A) 8 m/s
B) 6 m/s
C) 10 m/s
D) Not obtained from the data
Correct Answer: C
Solution :
Velocity of particle, \[v=\]rate of change of displacement \[ie,\] \[v=\frac{dr}{dt}\] 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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