• question_answer The point from where a ball is projected is taken as the origin of the coordinate axes. The$x$and y components of its displacement are given by$x=6t$and $y=8t-5{{t}^{2}}.$What is the velocity of projection? A) $~6\text{ }m{{s}^{-1}}$             B) $~8\text{ }m{{s}^{-1}}$C) $10\text{ }m{{s}^{-1}}$                D) $~14\text{ }m{{s}^{-1}}$

Solution :

$X=6t$and $y=8t-5{{t}^{2}}$ $\frac{dx}{dt}={{v}_{x}}=6$ $\frac{dv}{dt}={{v}_{y}}=8-10t,$$\therefore$${{\left. \frac{dy}{dt} \right|}_{t=0}}=8$ $\therefore$$v=\sqrt{{{v}_{x}}+v_{y}^{2}}=\sqrt{{{6}^{2}}+{{8}^{2}}}=10\,m/s$ Hence, the correction option is [c].

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