A particle A is projected from the ground with an initial velocity of $10 m s^{- 1}$ at an angle of 60° with horizontal. From what height h should an another particle B be projected horizontal with velocity $5 m s^{- 1}$ so that both the particles collide on the ground at point C if both are projected simultaneously? $(g = 10 m s^{- 2})$

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10 m

30 m

15 m

25 m

answer is C.

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Detailed Solution

Horizontal component of velocity of A is10 cos60° or $5 m s^{- 1}$ which is equal to the velocity of B in horizontal direction. They will collide at C if time of flight of the particles are equal $o r t_{A} = t_{B}$

$\frac{2 u \sin θ}{g} = \sqrt{\frac{2 h}{g}} (∵ h = \frac{1}{2} g t_{B}^{2})$

$o r h = \frac{2 u^{2} \sin^{2} θ}{g} = \frac{2 {(10)}^{2} {(\frac{\sqrt{3}}{2})}^{2}}{10} = 15 m$

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