Question

the froghopper philaenus spumarius is supposedly the best jumpre in the animal kingdom. to start a jump, this insect can accelerate at 4.00 km/s^2 over a distance of 2.00 mm as it straightens its specially adapted "jumping legs". Assume the acceleration is constant. Find the upward velocity with which the insect takes off

Answer 1

Use the formula
\[V^2-V_{0}^2=2 a x\]
a=4000 m/s^2 x=2 mm=0.002 m

Answer 2

at t=0 insect is at rest so V0=0

Answer 3

V is the take off velocity

Answer 4

soyou're taking the starting point as origin..mmhmm

Answer 5

so it'sjust v^2 = 2ax?

Answer 6

yes thats right

Answer 7

so v^2 = -2(4000)(0.002)?

Answer 8

why u put the negative sign?

Answer 9

lol woops nevermind that haha

Answer 10

so is that right?

Answer 11

yes i got v=4 m/s

Answer 12

4 m/s?

Answer 13

shouldnt it be 8

Answer 14

v^2=16 and v=4 ...

Answer 15

yup, i also got 4.

Answer 16

oops forgot to multiply by 2...thanks

Answer 17

then take the square root