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Standard XII

Physics

Multiplication with Vectors

A particle A ...

Question

A particle A is projected with speed $V_{A}$ from a point making an angle $60^{\circ}$ with horizontal. At the same instant, a second particle B is thrown vertically upwards from a point directly below the maximum height point of parabolic path of A, with velocity $V_{B}$ . If the two particles collide then the ratio of $\frac{V_{A}}{V_{B}}$ should be

1

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\frac{2}{\sqrt{3}}

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\frac{\sqrt{3}}{2}

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\sqrt{3}

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Solution

The correct option is B $\frac{2}{\sqrt{3}}$

To collide, their vertical component of velocity must be equal. $V_{A} s i n 60^{\circ} = V_{B} \Rightarrow V_{A} . \frac{\sqrt{3}}{2} = V_{B} \Rightarrow \frac{V_{A}}{V_{B}} = \frac{2}{\sqrt{3}}$

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Similar questions

Q. A particle is projected with speed

V_{A}

from a point making an angle

60^{0}

with horizontal. At the same instant, a second particle B is thrown vertically upward from point directly below the maximum height point of parabolic path of A with velocity

V_{B}

. If the two particles collide then the ratio of

V_{A} / V_{B}

should be:-

Q. A particle is projected with a speed

V

from a point O making an angle of

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with the vertical. At the same instant, a second particle is thrown vertically upwards with a speed

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from a point A. The two particles reach

H

, the highest point of the parabolic path of particle one simultaneously. Then ratio

V / v

Q. A particle is projected with a speed

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from a point O making an angle of

30^{\circ}

with the vertical. At the same instant, a second particle is thrown vertically upwards with a speed

v

from a point A. The two particles reach

H

, the highest point of the parabolic path of particle one simultaneously. Then ratio

V / v

Q. Two particles

A

and

B

are projected simultaneously from a point situated on a horizontal plane. The particle

A

is projected vertically up with a velocity

v_{A}

while the particle

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is projected up at an angle of

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with horizontal with a velocity

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(Take

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)

A particle P is projected with velocity $u_{1}$ at an angle of $30^{\circ}$ with the horizontal. Another particle Q is thrown vertically upwards with velocity $u_{2}$ from a point vertically below the highest point of path of P. The necessary condition for the two particles to collide at the highest point is

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The correct option is B 2√3 To collide, their vertical component of velocity must be equal. VA sin 60∘=VB⇒VA.√32=VB⇒VAVB=2√3

The correct option is B $\frac{2}{\sqrt{3}}$

To collide, their vertical component of velocity must be equal. $V_{A} s i n 60^{\circ} = V_{B} \Rightarrow V_{A} . \frac{\sqrt{3}}{2} = V_{B} \Rightarrow \frac{V_{A}}{V_{B}} = \frac{2}{\sqrt{3}}$