Question

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

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

Answer 1

Answer: (A)

The equation of trajectory is
$y=x \tan \theta-\frac{g x^{2}}{2 u^{2} \cos ^{2} \theta}$
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.
$g^{\prime}=\frac{9.8 \times 9}{25}=3.5 \,m s ^{-2}$