Two particles are seperates by a horizontal distance X and are launched with velocity U√3 at an angle of 30° and another with velocity U at an angle of 60° find the time after which the horizontal distance between them is 0

Question

Eshan · Answer

Horizontal component of velocity of first particle= $\sqrt{3}ucos30^{\circ}=\dfrac{3u}{2}$
Horizontal component of velocity of first particle= $ucos60^{\circ}=\dfrac{u}{2}$

Therefore the velocity of the two particles relative to one another is $\dfrac{3u}{2}-\dfrac{u}{2}=u$

Hence the time in which the horizontal distance between them is zero is $\dfrac{X}{u}$

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Two particles are seperates by a horizontal distance X and are launched with velocity U√3 at an angle of 30° and another with velocity U at an angle of 60° find the time after which the horizontal distance between them is 0

Two particles are seperates by a horizontal distance X and are launched with velocity U√3 at an angle of 30° and another with velocity U at an angle of 60° find the time after which the horizontal distance between them is 0

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Two particles are seperates by a horizontal distance X and are launched with velocity U√3 at an angle of 30° and another with velocity U at an angle of 60° find the time after which the horizontal distance between them is 0

Two particles are seperates by a horizontal distance X and are launched with velocity U√3 at an angle of 30° and another with velocity U at an angle of 60° find the time after which the horizontal distance between them is 0

1 Answers

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