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Equation of path of a projectile - (Concept)
$
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
| Exam | Chapter |
| JEE MAIN | Kinematics |
A projectile is given an initial velocity of , where
is along the ground and
is along the vertical. If
, the equation of its trajectory is
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The equation of trajectory of a projectile is given by where x and y are in metres and x is along horizontal and y is vertically upward and particles are projected from origin. Then which of the following option is incorrect
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A particle is moving with a velocity where K is a constant. The general equation for its path is:
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The trajectory of a projectile near the surface of the earth is given as . If it were launched at an angle
with speed
then
:
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The trajectory of a projectile in a vertical plane is , where
are constants and x and y are respectively the horizontal and vertical distances of the projectile from the point of projection. The angle of projection
and the maximum height attained H are respectively given by :
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If the initial velocity in horizontal direction of a projectile is unit vector $\hat{\mathrm{i}}$ and the equation of trajectory is $\mathrm{y}=5 \mathrm{x}(1-\mathrm{x})$. The y component vector of the initial velocity is $\qquad$.
(Take $\mathrm{g}=10 \mathrm{~m} / \mathrm{s}^2$ )
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A particle just clears a wall of height b at a distance a and strikes the ground at a distance c from the point of projection. The angle of projection is:
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When a body is projected from height h parallel to horizontal with velocity u then the equation of trajectory of the body is (taking upward direction as positive y-axis) -
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An aeroplane flying 490m above ground level at 100m/s, releases a block. After how much time (in s), it will hit the ground (g = 9.8 m/s2)?
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An aeroplane is flying horizontally with a velocity of 100m/s at a height of 500m when it is vertically at a point A on the ground, and a bomb is released from it. The bomb strikes the ground at point B. The distance AB is ____km (g = 9.8 m/s2)
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A projectile is given an initial velocity of (i+2j)m/s, where i is along the ground and j is along the vertical. If g=10m/s2, the equation of its trajectory is $y=a x-b x^2$. Find a+b.
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The equation of the trajectory of a projectile is given by . Find the maximum height in meters.
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Position of a particle moving in plane as function of time t is
. The equation of the trajectory of the particle is :
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A body of mass 10 kg is projected at an angle of $45^{\circ}$ with the horizontal. The trajectory of the body is observed to pass through a point $(20,10)$. If T is the time of flight, then its momentum vector, at time $\mathrm{t}=\frac{\mathrm{T}}{\sqrt{2}}$, is $\qquad$ [Take $g=10 \mathrm{~m} / \mathrm{s}^2$ ]
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A 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 :
| A. |
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| B. |
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| C. |
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The position of a particle is given by where t is in seconds and the coefficients have a proper unit for
to be in meters. The direction of v(t) at t=1s is
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The trajectory of the projectile in the vertical plane is , where a and b are constant and x and y are respectively the horizontal and vertical distances of the projectile from the point of projection. The projectile angle on the horizontal plane is:
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The trajectory of the projectile, projected from the ground is given by . Where
and
are measured in meters. The maximum height attained by the projectile will be.
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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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The equation of a projectile path is $y=4 x-11 x^2$; Then determine the angle of projection and initial velocity of projection. $\left(g=10 m s^{-1}\right)$
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A body is projected at an angle $\alpha$ to the horizontal so as to clear two waves of equal height h at a distance 2 h from each other, the horizontal range of a projectile is R. Then the value of $\operatorname{cot}\left(\frac{\alpha}{2}\right)$ in terms of R and h is given by:
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A small boy is throwing a ball towards a wall 6 m in front of him. He releases the ball and bounces from the ground. The ball bounces from the wall at a height of 3m, rebounds from the ground and reaches the boy's hand exactly at the point of release. Assuming the two bounces (one from the wall and the other from the ground) to be perfectly elastic, how far ahead of the boy did the ball bounce from the ground?
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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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Equation of path of a projectile is given by,
