Question 182: Motion in 2D & Vectors

Question

A projectile is fired from the surface of the earth with a velocity  of 5 m s
$$-$$1
and angle $$	heta $$ 
 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
)

Accepted Answer

The equation of trajectory is
$$y = x	an 	heta  - {{g{x^2}} \over {2{u^2}{{\cos }^2}	heta }}$$
where $$	heta $$ is the angle of projection and u is the
velocity with which projectile is projected.
For equal trajectories and for same angles of
projection,
$${g \over {{u^2}}}$$ = constant
According to the question, $${{9.8} \over {{5^2}}} = {{g'} \over {{3^2}}}$$
where g' is acceleration due to gravity on the planet.
$$g' = {{9.8 	imes 9} \over {25}} = 3.5\,m{s^{ - 2}}$$

Answer

3.5

Answer

5.9

Answer

16.3

Answer

110.8

Motion in 2D & Vectors

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