Question

A projectile is fired from the surface of the earth with a velocity of $5{\text{ms}}^{-1}$ and angle $\theta$ with the horizontal. Another projectile fired from another planet with a velocity of $3{\text{ms}}^{-1}$ 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 (in ${\text{ms}}^{-2}$) is

(Given $g=9.8{\text{ms}}^{-2}$)

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

Answer 1

The correct option is A $3.5$

If both the trajectories are identical, then
$H_e=H_p$ ($H$ is the maximum height of a projectile)
$\Rightarrow \frac{5^2{\sin }^2\theta }{2g_e}=\frac{3^2{\sin }^2\theta }{2g_p}$ ($\theta$ is same in both the case)
$\Rightarrow g_p=\frac{9}{25}{g}_e$
$=\frac{9}{25}\times 9.8=3.52{\text{m/s}}^2$
Hence, the correct answer is option (A).