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marysya [2.9K]
3 years ago
14

The froghopper Philaenus spumarius is supposedly the best jumper in the animal kingdom. To start a jump, this insect can acceler

ate 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.
Physics
1 answer:
Kay [80]3 years ago
5 0

Answer:

Part a)

v_f = 4 m/s

Part b)

t = 0.001 s

Part c)

d = 0.815 m

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

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