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The height y and distance x along the horizontal for a body projected in the xy-plane are given by $y = 8 t - 5 t^{2} a n d x = 6 t .$ The initial speed of projection is

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8 m/s

9 m/s

10 m/s

(10/3) m/s

answer is C.

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Detailed Solution

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

At t = 0. $v_{y} = 8 m / s a n d v_{x} = 6 m / s$

$∴ v = \sqrt{v_{x}^{2} + v_{y}^{2}} = 10 m / s$

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