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

The correct option is A $3.5$

Since both the projectiles follow the same path and same trajectories, their maximum height and range will be same.
Let the acceleration due to gravity on the planet be $g_p.$
Then, equation for maximum height is given as $H=\frac{u^2{\sin }^2\theta }{2g}$
$\Rightarrow \frac{5^2{\sin }^2\theta }{2g}=\frac{3^2{\sin }^2\theta }{2g_p}$
($\theta$ is same in both the case)
$\Rightarrow g_p=\frac{9}{25}g=\frac{9}{25}\times 9.8=3.52m/s^2$