A smooth square platform ABCD is moving towards right with a uniform speed v- At what angle - must a particle be projected from A with speed u so thatit strikes the point B -A- cos -1 v - ll B- cos -1u-vC- sin -1 v - N D- sin -1u- v

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Standard XII

Physics

Displacement in 2D Motion

A smooth squa...

Question

A smooth square platform ABCD is moving towards right with a uniform speed v. At what angle $θ$ must a particle be projected from A with speed u so that it strikes the point B?

$s i n^{- 1} (\frac{u}{v})$

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$c o s^{- 1} (\frac{v}{u})$

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$c o s^{- 1} (\frac{u}{v})$

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$s i n^{- 1} (\frac{v}{u})$

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Solution

The correct option is B
$c o s^{- 1} (\frac{v}{u})$

Particle will strike the point B if velocity of particle with respect to platform is along AB or component of its relative velocity along AD is zero i.e.,
$u c o s θ = v$ or $θ = c o s^{- 1} (\frac{v}{u})$

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A smooth square platform ABCD is moving towards right with a uniform speed v. At what angle θ must a particle be projected from A with speed u so that it strikes the point B?

The correct option is B cos−1(vu) Particle will strike the point B if velocity of particle with respect to platform is along AB or component of its relative velocity along AD is zero i.e., u cos θ=v or θ=cos−1(vu)

A smooth square platform ABCD is moving towards right with a uniform speed v. At what angle $θ$ must a particle be projected from A with speed u so that it strikes the point B?

The correct option is B
$c o s^{- 1} (\frac{v}{u})$

Particle will strike the point B if velocity of particle with respect to platform is along AB or component of its relative velocity along AD is zero i.e.,
$u c o s θ = v$ or $θ = c o s^{- 1} (\frac{v}{u})$