Question

A projectile is fired from the surface of the earth with a velocity of $5m{s}^{-1}$ and angle $\theta$ with the horizontal. Another projectile fired from another planet with a velocity of $3m{s}^{-1}$ 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 $g=9.8m{s}^{-2}$)

• A. 3.5
• B. 5.9
• C. 16.3
• D. 110.8

Answer 1

Answer: (A)

The equation of trajectory is
$y=x\tan \theta -\frac{g{x}^2}{2u^2{\cos }^2\theta }$
where $\theta$ is the angle of projection and $u$ is the velocity with which projectile is projected.
For equal trajectories and for same angles of projection,
$\frac{g}{u^2}=\text{ constant }$
As per question, $\frac{9.8}{5^2}=\frac{g^{\prime }}{3^2}$
where $g^{\prime }$ is acceleration due to gravity on the planet.
$g^{\prime }=\frac{9.8\times 9}{25}=3.5m{s}^{-2}$