The froghopper, a tiny insect, is a remarkable jumper. Suppose a colony of the little critters is raised on the Moon, where the acceleration due to gravity is only 1.62 m/s 2 , whereas gravity on Earth is g = 9.81 m/s 2 . If on Earth a froghopper's maximum jump height is h and its maximum horizontal jump range is R , what would its maximum jump height and range be on the Moon in terms of h and R ? Assume the froghopper's takeoff velocity is the same on the Moon and Earth.

hₘₒₒₙ = 6.05 h

Rₘₒₒₙ = 6.05 R

Explanation:

Let θ be the angle of jump.

Let h and R be maximum height and horizontal range attained on earth respectively.

Let hₘₒₒₙ and Rₘₒₒₙ be the maximum height and horizontal range on the moon respectively

The range for a projectile is given as

R = v₀(x)T = v₀ cos(θ) T

T = (2v₀ sinθ)/g

Range, R = (v₀ cos θ)(2v₀ sinθ)/g = v₀²(2sinθcosθ)/g = v₀² (sin2θ)/g

The maximum range occurs at θ = 45°

Maximum range R = v₀²/g = v₀²/9.8 = 0.102v₀²

On the moon, g = 1.62 m/s²

Maximum range, Rₘₒₒₙ = v₀²/gₘₒₒₙ = v₀²/1.62 = 0.617v₀²

Rₘₒₒₙ = 6.05 R

Maximum Height of a projectile is given as = (v₀² Sin²θ)/2g

θ = 45°; sin 45° = (√2)/2; sin²45° = 2/4 = 1/2

h = v₀²(1/2)/2g = v₀²/4g

On earth, g = 9.8 m/s²

h = v₀²/(4×9.8) = v₀²/39.2 = 0.0255v₀²

On the moon, gₘₒₒₙ = 1.62 m/s²

hₘₒₒₙ = v₀²/(4×1.62) = v₀²/6.48 = 0.154v₀²

hₘₒₒₙ = 6.05 h

fgravity=10x10=100n case 1

fgravity=30x20=600n case 2

6 times more

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