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    Equation Of Path Of A Projectile - Practice Questions & MCQ

    Edited By admin | Updated on Sep 18, 2023 18:34 AM | #JEE Main

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    • 18 Questions around this concept.

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    A projectile is given an initial velocity of \left ( \hat{i}+2\hat{j} \right )m/s , where \hat{i} is along the ground and \hat{j} is along the vertical. If g = 10m/s2, the equation of its trajectory is :

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    An initial velocity of $(2 \hat{i}+\sqrt{3} \hat{j}) \mathrm{m} / \mathrm{s}$ is given to a projectile, which $\hat{i}$ is along the ground and $\hat{j}$ is along the vertical. The equation of its trajectory is if $g=10 \mathrm{~m} / \mathrm{s}^2$.

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    A particle is moving according to the equation $y=\left(\sqrt{3} X-2 X^2\right)$. Then the horizontal range of project is 

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    Concepts Covered - 1

    Equation of path of a projectile

        

    $
    y=x \tan \theta-\frac{g x^2}{2 u^2 \cos ^2 \theta}
    $

    or-

    $
    y=x \tan \theta\left(1-\frac{x}{R}\right)
    $


    Where, R is the horizontal range of the projectile.
    It is equation of parabola, So the trajectory path of the projectile is parabolic in nature
    $g \rightarrow \quad$ Acceleration due to gravity
    $u \rightarrow$ initial velocity
    $\theta=$ Angle of projection

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    Equation of path of a projectile

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    Equation Of Path Of A Projectile Current Topic

    Kinematics Terminologies
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    Scalars And Vectors
    Vector Addition And Vector Subtraction
    Multiplication Of Vectors
    Distance And Displacement
    Speed And Velocity
    Acceleration
    Kinematics Graphs
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