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The minimum and maximum velocities of a projectile are $10 {ms}^{- 1}$ and $20 {ms}^{- 1}$ respectively. The horizontal range and maximum height are respectively $(g = 10 m s^{- 2})$

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$10 \sqrt{3} m and 10 m$

$10 \sqrt{3} m and 20 m$

$20 \sqrt{3} m and 15 m$

20m and 15m

answer is B.

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Detailed Solution

ucos $θ$ = 10

and u = 20

$\Rightarrow \cos θ = \frac{1}{2}$

$\Rightarrow θ = 60 °$

$Horizontal range = R = \frac{u^{2} \sin 2 ϕ}{g} = \frac{20^{2} \sin 120^{\circ}}{10} R = 20 \sqrt{3} m Maximum height = H = \frac{u^{2} \sin^{2} ϕ}{2 g} = \frac{20^{2} \times \frac{3}{4}}{2 \times 10 g} H = 15 m$

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