Question

The froghopper Philaenus spumarius is supposedly the best jumper in the animal kingdom. To start a jump, this insect can accelerate at 4.00 km/s2 over a distance of 2.0 mm as it straightens its specially designed "jumping legs."

(a) Find the upward velocity with which the insect takes off.
(b) In what time interval does it reach this velocity?
(c) How high would the insect jump if air resistance were negligible? The actual height it reaches is about 70 cm, so air resistance must be a noticeable force on the leaping froghopper.

Answer 1

Part a)

`v_f = 4 m/s`

Part b)

`t = 0.001 s`

Part c)

`d = 0.815 m`

Step-by-step explanation:

Part a)

As we know that initially the grass hopper is at rest at the ground position

Now the acceleration is given as

`a = 4000 m/s^2`

distance of the legs that it stretched is given as

`s = 2.0 mm`

so we have

`v_f^2 - v_i^2 = 2 a d`

`v_f^2 - 0 = 2(4000)(0.002)`

`v_f = 4 m/s`

Part b)

time taken to reach this speed is given as

`v_f - v_i = at`

`4 - 0 = 4000 t`

`t = 0.001 s`

Part c)

as the grass hopper reach the maximum height its final speed would be zero

so we will have

`v_f^2 - v_i^2 = 2 a d`

`0 - 4^2 = 2(-9.81) d`

`d = 0.815 m`