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Trajectory of particle in a projectile m...

Trajectory of particle in a projectile motion is given as: `y=x-(x^(2)//80)` (x and y are in meters). Take `g=10m//s^(2)`.

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Trajectory of particle in a projectile motion is given as y=x- x^(2)/80 . Here, x and y are in meters. For this projectile motion, match the following with g=10 m//s^(2) . {:(,"Column-I",,"Column-II"),((A),"Angle of projection (in degrees)",(P),20),((B),"Angle of velocity with horizontal after 4s (in degrees)",(Q),80),((C),"Maximum height (in meters)",(R),45),((D),"Horizontal range (in meters)",(S),30),(,,(T),60):}

Trajectory of a particle in a projectile motion is given as y=x-(x^(2))/80 . Here x and y are in metre. For this projectile motion the following with g=10m//s^(2) .

Knowledge Check

  • Equation of trajectory of a projectile is given by y = -x^(2) + 10x where x and y are in meters and x is along horizontal and y is vericall y upward and particle is projeted from origin. Then : (g = 10) m//s^(2)

    A
    initial velocity of particle is `sqrt(505) m//s`
    B
    horizontal range is `10 m`
    C
    maximum height is `25 m`
    D
    angle of projection with horizontal is `tan^(-1)(5)`.

  • The equation of trajectory followed by a projectile is given by y =x-(x^(2))/(2) . At any instant, its x-coordinate is given by x-(t)/(2) . At what time (in sec.) does the projectile reach y = (1)/(2)m ?

    A
    2
    B
    4
    C
    Never
    D
    3

  • The trajectory of a projectile in a vertical plane is y=sqrt(3)x-2x^(2) . [g=10 m//s^(2)] Range OA is :-

    A
    `sqrt(3)/2`
    B
    `sqrt(3)/4`
    C
    `sqrt(3)`
    D
    `3/8`

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    Trajectory of a particle in a projectile motion is given by y=x-x^(2)/80 where x and y are in meters. Match the column-1 and column-2. {:(,"Column-I",,"Column-II"),("(A)","x coordinate at height of 15 m",,"(P) 20 m"),("(B)",underset("point of projection at x= 100 m")"vertical distnce of particle from",,"(Q) 80 m"),("(C)","Horizontal range",,"(R) 60 m"),(,,,"(S) 25 m"):}

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