Q. A ball is thrown with a velocity of u making an angle θ with the horizontal. Its velocity vector normal to initial vector (u) after a time interval of
(a) $\displaystyle \frac{u sin\theta}{g}$
(b) $ \displaystyle \frac{u }{g cos\theta}$
(c) $ \displaystyle \frac{u }{g sin\theta}$
(d) $ \displaystyle \frac{u cos\theta}{g}$
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Ans: (c)
Sol: $ \displaystyle \vec{u} = ucos\theta \hat{i} + usin\theta \hat{j} $
$ \displaystyle \vec{v} = v_x\hat{i} + v_y \hat{j} $
$ \displaystyle \vec{v} = ucos\theta \hat{i} + (usin\theta – gt) \hat{j} $
Since u & v are perpendicular then
$ \displaystyle \vec{u}.\vec{v} = 0 $
$ \displaystyle t = \frac{u}{g sin\theta} $