Equation Of Path Of A Projectile - Practice Questions & MCQ
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18 Questions around this concept.
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EASYA projectile is given an initial velocity of , where
is along the ground and
is along the vertical. If g = 10m/s2, the equation of its trajectory is :
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 $g=10 \mathrm{~m} / \mathrm{s}^2$.
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
$
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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