The froghopper, Philaenus spumarius, holds the world record for insect jumps. When leaping at an
angle of 58.0 degrees above the horizontal, some of the tiny critters have reached a maximum height
of 58.7 cm above the level ground. (See Nature, Vol. 424, 31 July 2003, p. 509.)
(a) What was the take-off speed for such a leap?
(b) What horizontal distance did the froghopper cover for this world record leap?

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  1. h = Vi t - 4.9 t^2
    .587 = Vi t - .5*9.81 t^2

    at top is Vi/9.81
    where 0 = Vi - g t

    .587 = Vi^2/9.81 - .5 * 9.81 (Vi^2/9.81^2)

    .587 = .5 * Vi^2/9.81
    Vi = 3.39 m/s vertical up

    Vi = speed * sin 58
    speed = 4.00 m/s part a

    u = 4.00 cos 58 = 2.12 m/s

    time in air = 2 * rise time
    rise time = Vi/g
    time in air = 2 * 3.39/9.81
    = .691 seconds in air
    range = u * .691
    = 2.12 * .691
    = 1.47 meters
    wow, that is impressive if correct.

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  2. why is a=4.9

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  3. 4.9 is (1/2)*a, a = gravity which is -9.8

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