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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

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Solution

The correct option is **D** 2(v+ucosθ)usinθg

A shell is fired upward with velocity u , inclined at angle θ

with the horizontal

The horizontal component for shell will be

=ucosθ

Velocity of car is give as ν

Hence total horizontal velocity component, v+ucosθ⟶1

Let the time taken by the projectile motion be t, when the shells reaches ground, the vertical displacement is zero. Also from second equation of motion

We know that,

y=utsinθ−12gt2

We are considering vertical component of velocity of the shell, putting y=0 in second equation of motion we get

t=2usinθg⟶2

Range is horizontal distance traveled which will be the product of horizontal component of velocity s time

∴ Range=(v+ucosθ)t

Substituting values of t from equation 2

Range=2(v+ucosθ)usinθg

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