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Motion in A Plane question

2014 · Q140
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Motion in A Plane question

2014 · Q140

NEETPhysicsMotion in A PlaneMCQ+4 / −1
A projectile is fired from the surface of the earth with a velocity of 5 m s−-−1 and angle θ\thetaθ with the horizontal. Another projectile fired from another planet with a velocity of 3 m s−-−1 at the same angle follows a trajectory which is identical with the trajectory of the projectile fired from the earth. The value of the acceleration due to gravity on the planet is (in m s−-−2) is
(Given g = 9.8 m s−-−2)
  1. A
    3.5
  2. B
    5.9
  3. C
    16.3
  4. D
    110.8
View written solutionFree

Correct answer: A

The equation of trajectory is

y=xtan⁡θ−gx22u2cos⁡2θy = x\tan \theta - {{g{x^2}} \over {2{u^2}{{\cos }^2}\theta }}y=xtanθ−2u2cos2θgx2​

where θ\theta θ is the angle of projection and u is the velocity with which projectile is projected. For equal trajectories and for same angles of projection,

gu2{g \over {{u^2}}}u2g​ = constant

According to the question, 9.852=g′32{{9.8} \over {{5^2}}} = {{g'} \over {{3^2}}}529.8​=32g′​

where g' is acceleration due to gravity on the planet.

g′=9.8×925=3.5 ms−2g' = {{9.8 \times 9} \over {25}} = 3.5\,m{s^{ - 2}}g′=259.8×9​=3.5ms−2

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