A projectile is fired from the surface of the earth with a velocity of 5 m s⁻¹ and angle θ 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 eart…

Question

A projectile is fired from the surface of the earth with a velocity of 5 m s⁻¹ and angle θ 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)

Accepted Answer

(1) 3.5. Given: same launch angle, speeds 5 m s⁻¹ on earth and 3 m s⁻¹ on the planet, identical trajectories. Idea: the path equation is y=x tan θ-(gx²)/(2u² cos²θ), so identical paths at the same angle require g/(u²) to be the same. (gₑ)/(uₑ²)=(gₚ)/(uₚ²). (9.8)/(25)=(gₚ)/9. gₚ=(9.8×9)/(25)=(88.2)/(25)=3.5 m s⁻².

Step-by-step solution

Why the other options are wrong