The click beetle can project itself vertically with an acceleration of about 400

Question

The click beetle can project itself vertically with an acceleration of about 400g (an order of magnitude more than a human could survive!). The beetle jumps by "unfolding" its 0.62-cm long legs.

a) How high can the beetle jump? Answer in meters.

b)How long is the beetle in the air? (Assume constant acceleration while in contact with the ground and neglect air resistance.) In seconds.

I don't even know where to start, with what kinematic equation? No matter where I start I seem to get a wrong answer. I realize it is a two part velocity problem. And once I get distance it is pretty easy to relay time back with the kinematic equations.

Explanation / Answer

Yes, this is a little different than your typical problem, you could say it has a new twist, but it is a click beetle, not a twist beetle. (sorry) Once the beetle gets off the ground, the game is over and it is just a regular projectile like we are used to, so the big acceleration is only when the beetle is kicking off. We will use the big acceleration to find the speed it has when it takes off. This speed is the final speed in the very first part of the problem, but once we have it, it will be the starting speed for parts a) and b). Launch: We assume constant acceleration. The acceleration takes place while the beetle unfolds its 0.62 cm legs. It starts out lying on its back (I think) with zero velocity. So we have a, Vo, x-xo. Do we know a formula that will give us Vf? Yes, it's that wonderful formula with the velocities squared. Vf^2 = Vo^2 + 2 a (x-xo) It is already solved for the variable we need, so no algebra required. Vo is zero. 0.62 cm is 0.0062 m Vf^2 = 2 400 (9.8) 0.0062 = 48.6 m^2/s^2 Vf = 6.97m/s (a) how high can the beetle jump? If the beetle jumps straight up, it's starting velocity is 6.97 m/s At its highest point its upward speed is 0 m/s Once the beetle is airborne, a = - 9.8 m/s^2 Again, we want the formula with the squared velocities Vf^2 = Vo^2 + 2 a (x-xo) Vf = 0 Solve for (x-xo) x-xo = -Vo^2/2 a = -(6.97m/s)^2/[(-9.8 m/s^2)*2) = 2.48 m does this seem right? Yes, I have seen click beetles shoot up about a meter, so they could maybe go this high. (b) how long is it in the air. Try to get a feel for how you could quickly estimate what an answer should be. If it is going up at 7 m/s, it slows down 9.8 m/s each second, so it will be going up at 0 m/s before one second has gone by and then it will take the same time to fall back down. It starts and lands at the same height, so (x-xo)=0 x-xo = Vot + 1/2 a t^2 hmmm... that seems a little tough. Let's chop the problem in half starting at the highest point, and calling that Vo. We do this since the vertical speed at the top is zero so it is easier. t^2 = 2(x-xo)g = 2*(-2.48m)/(-9.8m/s^2) = 0.506122 t = 0.71s is the time it falls from its highest point to the ground. It takes the same time to go up to the highest point. Time in the air = 1.4 s

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