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A particle is projected vertically upwar...

A particle is projected vertically upwards and t second after another particle is projected upwards with the same initial velocity. Prove that the particles will meet after a lapse of `((t)/(2)+(u)/(g))` time from the instant of projection of the first particle.

Text Solution

AI Generated Solution

To solve the problem, we need to analyze the motion of both particles and find the time at which they meet after the first particle is projected. Let's denote the following: - \( u \): initial velocity of both particles - \( g \): acceleration due to gravity - \( t \): time difference between the projections of the two particles ### Step 1: Analyze the motion of the first particle ...
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Knowledge Check

  • A particle is projected vertically upwards in vacuum with a speed u .

    A
    When it rises to half its maximum height, its speed becomes `u//2`.
    B
    When it rises to half its maximum height, its speed becomes `u//sqrt(2)`.
    C
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    D
    The time taken to rise to three`-` fourth of its maximum height is half the time taken to reach its maximum height.

  • A body A is projected upwards with a velocity of 90m//s . The second body B is projected upwards with the same initial velocity but after 4 sec. Both the bodies will meet after

    A
    6sec
    B
    8sec
    C
    10sec
    D
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  • A particle P is projected upwards with 80m//s . One second later another particle Q is projected with initial velocity 70m//s . Before either of the particle srikes the ground (g=10m//s^(2))

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    B
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    C
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    D
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