A projectile thrown with a speed v at an angle - has a range R on the surface of earth- For same v and - its range on the surface of moon will be

The correct option is A R=u^2 sin 2 θ/g ∴ R ∝ 1/g R_Moon/R_Earth=g_Earth/g_Moon=6 [ ∵ g_Moon=1/6 g_Earth ] ∴ R_Moon = 6 R_Earth=6 R ...

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Byju's Answer

Standard XII

Physics

Range

A projectile ...

Question

A projectile thrown with a speed v at an angle $θ$ has a range R on the surface of earth. For same v and $θ$ , its range on the surface of moon will be

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Solution

The correct option is B

$R = \frac{u^{2} s i n 2 θ}{g} ∴ R \propto \frac{1}{g} \frac{R_{M o o n}}{R_{E a r t h}} = \frac{g_{E a r t h}}{g_{M o o n}} = 6 [∵ g_{M o o n} = \frac{1}{6} g_{E a r t h}] ∴ R_{M o o n} = 6 R_{E a r t h} = 6 R$

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A projectile thrown with a speed v at an angle θ has a range R on the surface of earth. For same v and θ, its range on the surface of moon will be

The correct option is B R=u2 sin 2 θg ∴ R ∝ 1gRMoonREarth=gEarthgMoon=6 [∵ gMoon=16 gEarth]∴ RMoon=6 REarth=6 R

A projectile thrown with a speed v at an angle $θ$ has a range R on the surface of earth. For same v and $θ$ , its range on the surface of moon will be

The correct option is B

$R = \frac{u^{2} s i n 2 θ}{g} ∴ R \propto \frac{1}{g} \frac{R_{M o o n}}{R_{E a r t h}} = \frac{g_{E a r t h}}{g_{M o o n}} = 6 [∵ g_{M o o n} = \frac{1}{6} g_{E a r t h}] ∴ R_{M o o n} = 6 R_{E a r t h} = 6 R$