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 $x = 6 t$ and $y = 8 t - 5 t^{2}$ . What is the velocity of projection?

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$8 m s^{- 1}$

$14 m s^{- 1}$

$10 m s^{- 1}$

$6 m s^{- 1}$

answer is C.

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Detailed Solution

$v_{x} = \frac{d x}{d t} = 6 a n d v_{y} = \frac{d y}{d t} = 8 - 10 t$

Putting $t = 0$ (since we have to find initial velocity)

$v_{y} = 8 - 10 \times 0 = 8$

$v = \sqrt{v_{x}^{2} + v_{y}^{2}} = \sqrt{6^{2} + 8^{2}} = 10 m s^{- 1}$

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