a projectile is fired from the surface of the earth with a velocity of `5ms^(-1)` and angle `theta` with the horizontal. Another projectile fired from another planet with a velocity of `3ms^(-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))`

Options

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

Answer

A

Explanation

(a): The acceleration due to gravity will be different for different planets .
So, `R=(u^(2))/(g) sin 2 theta`
`rArr gpropu^(2)`
`[R and theta`are same for both projectiles]
`(g_("earth"))/(g_("planet"))=(u_("earth")^(2))/(u_("planet")^(2))`
`g_("planet")=(gxxu_("planet")^(2))/(u_("earth")^(2))`
`=(9.8xx(3)^(2))/((5)^(2))`
`g_("planet")=3.528 m//s^(2)`