Question

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?

Answer 1

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

but t 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
so
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
so
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.

Answer 2

why is a=4.9

Answer 3

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