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The horizontal and vertical displacements of a projectile, projected obliquely on the surface of a planet as a function of time, are given by $x = 10 t m and y = (12 t - 4 t^{2}) m$ . The horizontal range of the projectile is

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zero

120 m

30 m

20.6 m

answer is A.

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Detailed Solution

a) $x = 10 t$

$x = u \cos θ t$

$\Rightarrow u \cos θ = 10 m / s$

b) $y = 12 t - 4 t^{2}$

$y = u \sin θ t - \frac{1}{2} {gt}^{2}$

$u \sin θ = 12 m / s$

$\frac{g}{2} = 4$

$g = 8$

$R = \frac{2 (U \cos θ) (u \sin θ)}{g}$

$\frac{2 [10] [12]}{8} = 30 m$

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