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

ill do it for 100, also this isnt 80 its 40

neither accurate nor precise.