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?

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

speed of light in vacuum is 300,000 km/s

speed of light in other mediums is given by the expression v = c/n, where c is the speed of light in vacuum and n is the refractive index of medium.

for the first medium v = 200,000 km/s

  so, refractive index of medium = 300000/200000 = 1.5

    1.5 refractive index is usually for glass. so first medium is glass.

for the second medium v = 250,00 km/s

so, refractive index of medium = 300000/250000 = 1.2

    1.2 refractive index is usually for water. so second medium is water.

for the third medium v = 0 km/s

so, refractive index of medium = 300000/0 = ∞

    ∞ refractive index is usually for opaque mediums. so third medium is opaque medium.