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The height y and the distance x along the horizontal plane of a projectile on a certain planet (with no surrounding atmosphere) are given by y = (8t - 5t²) m and x = 6t m, where t is in seconds. The velocity with which the projectile is projected at t = 0 is

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10 ms-1

Not obtainable from the data

8 ms^-1

6 ms-1

answer is A.

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Detailed Solution

$x = 6 t, u_{x} = \frac{dx}{dt} = 6 y = 8 t - 5 t^{2}, v_{y} = \frac{dy}{dt} = 8 - 10 t v_{y} (t = 0) = 8 {ms}^{- 1} u = \sqrt{u_{x}^{2} + u_{y}^{2}} = \sqrt{6^{2} + 8^{2}} = 10 {ms}^{- 1}$

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