Suppose an object is thrown with initial speed of 10 `ms^(-1)` at an angle `pi//4` with the horizontal, what is the range-covered? Suppose the same object is thrown similarly in the moon, will there be any change in the range? If yes, what is the change? (The acceleration due to gravity in the moon `g_("moon")=1//6 g`)

Explanation

In projectile motion, the range of particle is given by,
`R=(u^(2)sin2theta)/g`
`theta=pi/4u=v_(0)=10ms^(-1)`
`therefore R_("earth")=((10)^(2)sin""pi/2)/9.8=100/9.8`
`R_("earth")=10.20 m` (Approximately 10 m)
If the same object is thrown in the Moon, the range will increase because in the Moon, the acceleration due to gravity is smaller than g on Earth,
`g_("moon")=g/6`
`R_("moon")=(u^(2)sin2theta)/(g_("moon"))=(v_(0)""^(2)sin2theta)/(g/6)`
`therefore R_("moon")=6R_("earth")`
`R_("moon")=6 times 10.20=61.20m`
(Approximately 60 m)
The range attained on the Moon is approximately six times that on Earth.