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Question

# A car is moving horizontally with velocity v. A shell is fired upward with velocity u inclined at angle θ with the horizontal. The horizontal range of the shell related to ground is

A
2u2vg
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B
2v2ug
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C
2u2v2g
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D
2(v+ucosθ)usinθg
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Solution

## The correct option is D 2(v+ucosθ)usinθgA shell is fired upward with velocity u , inclined at angle θ with the horizontalThe horizontal component for shell will be=ucosθVelocity of car is give as νHence total horizontal velocity component, v+ucosθ⟶1Let the time taken by the projectile motion be t, when the shells reaches ground, the vertical displacement is zero. Also from second equation of motionWe know that,y=utsinθ−12gt2We are considering vertical component of velocity of the shell, putting y=0 in second equation of motion we gett=2usinθg⟶2Range is horizontal distance traveled which will be the product of horizontal component of velocity s time ∴ Range=(v+ucosθ)tSubstituting values of t from equation 2Range=2(v+ucosθ)usinθg

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