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Question: Assume that the acceleration due to gravity on the surface of the moon is 0.2 times the acceleration...

Assume that the acceleration due to gravity on the surface of the moon is 0.2 times the acceleration due to gravity on the surface of the earth. If ReR _ { e } is the maximum range of a projectile on the earth’s surface, what is the maximum range on the surface of the moon for the same velocity of projection?

A

0.2 R _ { e }

B

2Re2 R _ { e }

C

0.5Re0.5 R _ { e }

D

5Re5 R _ { e }

Answer

5Re5 R _ { e }

Explanation

Solution

Range of projectile R=u2sin⁡2θgR = \frac { u ^ { 2 } \sin 2 \theta } { g }

if u and θ are constant then R∝1gR \propto \frac { 1 } { g }

RmRe=gegm⇒RmRe=10.2⇒Rm=Re0.2\frac { R _ { m } } { R _ { e } } = \frac { g _ { e } } { g _ { m } } \Rightarrow \frac { R _ { m } } { R _ { e } } = \frac { 1 } { 0.2 } \Rightarrow R _ { m } = \frac { R _ { e } } { 0.2 }⇒ Rm=5ReR _ { m } = 5 R _ { e }