The froghopper, Philaenus spumarius, holds the world record for insect jumps. When leaping at an angle of 58.0∘ above the horizontal, some of the tiny critters have reached a maximum height of 58.7 cm above the level ground. (a) What was the takeoff speed for such a leap? (b) What horizontal distance did the froghopper cover for this world-record leap?

Question

Physics

Talon Arnold

20 May, 18:48

The froghopper, Philaenus spumarius, holds the world record for insect jumps. When leaping at an angle of 58.0∘ above the horizontal, some of the tiny critters have reached a maximum height of 58.7 cm above the level ground. (a) What was the takeoff speed for such a leap? (b) What horizontal distance did the froghopper cover for this world-record leap?

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Answers (1)

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Shane Adams · Answer 1

a) v₀ = 4 m/s

b) Xmax = 1.467 m

Explanation:

Given

∅ = 58°

ymax = 58.7 cm = 0.587 m

a) v₀ = ?

We use the formula

ymax = (v₀*Sin ∅) ² / (2g)

⇒ v₀ = √ (2g*ymax) / Sin ∅ = √ (2 * (9.81 m/s²) * (0.587 m)) / Sin 58°

⇒ v₀ = 4 m/s

b) We apply the equation

Xmax = v₀²*Sin (2∅) / g

⇒ Xmax = (4 m/s) ²*Sin (2*58°) / (9.81 m/s²) = 1.467 m