The height y and the distance x along the horizontal plane bf the projectile on a certain planet (with no surrounding atmosphere) are given by
$y = (8 t - 5 t^{2}) metre and x = 6 t metre$
where t is in second. The velocity with which the projectile is projected is

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8 m/sec

6 m/sec

10 m/sec

Not obtainable from the data given

answer is C.

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Detailed Solution

$v_{y} = \frac{dy}{dt} = 8 - 10 t = 8$ when t = 0 at the time of projection

$\begin{array}{l} v_{x} = \frac{dx}{dt} = 6 \\ v = \sqrt{(v_{x}^{2} + v_{y}^{2})} = \sqrt{\{(8)^{2} + (6)^{2}\}} \\ = 10 m / \sec \end{array}$

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