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 `3 m*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)`) (given `g=9.8 m*s^(-2)`)

Options

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

Answer

A

Explanation

Since trajectories of both cases are identical,
`R_(max)=R_(max)^.[ "Here" R_(max)^.` is the horizontal range in the second case ]
`therefore R_(max) =(u_1^2)/(g) sin2 theta and R_(max)^. =(u_2^2)/(g^.) sin 2 theta`
So, `(u_1^2)/(g) sin 2theta=(u_2^2)/(g^.) sin 2theta`
`therefore g^.=(u_2^2g)/(u_1^2) =(3^2xx9.8)/(5^2) =3.5`