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Equation Of Path Of A Projectile MCQ - Practice Questions with Answers

Edited By admin | Updated on Sep 25, 2023 25:23 PM | #NEET

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  • 17 Questions around this concept.

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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:

An object is projected with a velocity of 20 \mathrm{~m} / \mathrm{s} making an angle of  45^{\circ} with horizontal The equation for the trajectory is \mathrm{h=A x-B x^2} is height  \mathrm{x }  is horizontal distance, \mathrm{A}and \mathrm{B } are constant, the ratio \mathrm{A: B } is

A body of mass 20 \mathrm{~kg} is being Projected with an angle of 30^{\circ} with the vertical. The trajectory of body is observed at point  (30,20) If the ' \mathrm{ T} ', is time of fight Then its momentum vector, at time \mathrm{ t=\frac{T}{\sqrt{3}}} is

If the initial velocity in the Horizontal direction of a projectile is unit vector i^ and the equation of trajectory is y=10x(1x). Find the y -component vector of initial velocity (g=10 m/sec2)

Assertion: For x=10m;y=12x610x2, this projectile motion the equation will reach the maximum height

Reason: Maximum height depends upon the initial velocity of the projectile.

 

For a parabolic path of projection y=xgx22u2cos2Θ. The angle of projection is 


 

The equation of projectile is given as y=xgx210. Then the maximum range of the projectile is ( x and y are in meters)

 

Concepts Covered - 1

Equation of path of a projectile

y=xtanθgx22u2cos2θ


It is equation of parabola, So the trajectory/path of the projectile is parabolic in nature.
g Acceleration due to gravity
u initial velocity
θ= Angle of projection

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Equation of path of a projectile

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