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)

Suppose an object is thrown with initial speed of 10 ms^(-1) at an ang

1.	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)
Answer» SOLUTION :In projectile motion, the RANGE of particle is given by, `R=(U^(2)SIN2THETA)/(g)` `theta=pi//4u=v_(o)=10ms^(-1)` `R_("earth")=((10)^(2)sinne//2)/(9.8)=100//9.8` `R_("earth")=10.20m` (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)` `:.R_("moon")=6R_("earth")` `R_("moon")=6xx10.20=61.20m`(Approximately 60 m) The range attained on the Moon is approximately SIX times that on Earth.