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A body of mass 2kg is projected from the ground with a velocity 20ms^–1 at an angle 30° with the vertical. If t₁ is the time in seconds at which the body is projected and t₂ is the time in seconds at which it reaches the ground, the change in momentum in kg ms^–1 during the time (t₂ – t₁) is

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50 $\sqrt{3}$

40 $\sqrt{3}$

answer is B.

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Detailed Solution

t₂–t₁ is time of flight.

The change in momentum during the time of flight = 2musin $θ$
$= 2 \times 2 \times 20 \times \frac{\sqrt{3}}{2} = 40 \sqrt{3}$ $(∵ Angle with horizontal θ = 90 - 30 = 60^{0})$

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