Certain insects can achieve seemingly impossible accelerations while jumping. the click beetle accelerates at an astonishing 400

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Certain insects can achieve seemingly impossible accelerations while jumping. the click beetle accelerates at an astonishing 400

g over a distance of 0.52 cm as it rapidly bends its thorax, making the "click" that gives it its name. part a assuming the beetle jumps straight up, at what speed does it leave the ground? part b how much time is required for the beetle to reach this speed? part c ignoring air resistance, how high would it go?

Physics

1 answer:

inna [3.1K] · Answer 1 · 2026-02-22T07:07:58+03:00

(a) The beetle's launch speed is 6.4 m/s.

(b) The duration to reach the launch speed is 1.63 ms.

(c) The maximum altitude attained by the beetle is 2.1 m.

(a) Starting from rest, the beetle accelerates upward at 400 g, reaching the launch speed over a distance of 0.53 cm, where g represents gravitational acceleration.

Using the motion equation,

$v^2=u^2+2as$

Here, u indicates the initial velocity of the beetle, v the final velocity, a the beetle's acceleration, and s the distance covered.

The values used are 0 m/s for u, 400 g for a, 9.8 m/s² for g, and 0.52 × 10⁻² m for s.

$v^2=u^2+2as\\ = (0 m/s)^2+2 (400)(9.8 m/s^2)(0.52*10^-^2 m)\\ =40.768 (m/s)^2\\ v=6.385 m/s$

The resulting launch speed of the beetle is 6.4 m/s.

(b) To calculate the time t needed for the beetle's upward launch, use the motion equation:

$v=u+at$

Substituting 0 m/s for u, 400 g for a, 9.8 m/s² for g, and 6.385 m/s for v provides:

$v=u+at\\ 6.385 m/s = (0 m/s) +400(9.8 m/s^2)t\\ t = \frac{6.385 m/s}{3920 m/s^2} = 1.63*10^-^3s=1.63 ms$

The beetle takes 1.62 ms to launch itself.

(c) Upon launching, the beetle is subject to gravitational force, drawing it downward with an acceleration matching gravity g. Its ascent speed decreases until it reaches zero at the peak height.

Use the equation of motion:

$v^2=u^2+2as$

Inputting 6.385 m/s for u, -9.8 m/s² for g, and 0 m/s for v yields:

$v^2=u^2+2as\\ (0m/s)^2=(6.385 m/s)^2+2(-9.8m/s^2)s\\ s=\frac{(6.385 m/s)^2}{2(9.8m/s^2)} =2.08 m$

The beetle reaches a height of 2.1 m.