How Long Does a Cricket Take to Jump Portions of Its Full Leap?

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The discussion centers on calculating the time it takes for a cricket to jump 7 cm of its maximum 21 cm leap. Initial assumptions include a jump height of 21 cm, gravitational acceleration of 9.80 m/s², and an initial speed of zero. A calculation suggests it takes approximately 0.069 seconds to reach 7 cm, based on the assumption that time is proportional to height. However, participants argue that the cricket's initial speed is not zero and that time does not scale linearly with height due to varying acceleration. The conversation emphasizes the need to consider the cricket's changing speed and provides guidance on using kinematic equations to solve the problem effectively.
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Heres the question: Some crickets can jump 21 cm, how long does it take the cricket to jump 7cm of that 21cm?
 
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How do you suppose to solve this question?
 
hmmmm...well..we need to fulfill 3 variables in order to solve this or else it won't work.

h = 21cm which is 0.21m
g = 9.80m/s2
v-initial = 0 (I am assuming cricket is at rest before it jumps)

Using this equation:

h = v-init (t) + 1/2gt2(squared)

t2 = 0.043
t = 0.207s

7cm of 21cm is 1/3 of the height, so since it takes 0.207s for the cricket to get up to 21 cm (0.21m) we want 1/3 of the time.

1/3 X 0.207s = 0.069s to reach 7 cm.


I don't know if this is right. So don't go by my answer 100%. Maybe someone can verify.
 
Hmm,

The initial speed is not 0, as the cricket wouldn't go up for starters. In order to solve this question, you should make abstraction (ie forget about) the acceleration of the cricket, that is the time it takes for him to go from rest to full take-off speed.

It does not take 1/3rd of the time to travel to 1/3rd of the height, as in the beginning the cricket is going very fast, and is gradually loosing speed, therefore requiring less time to do the first part of the jump than for the last part of it.

So, how to tackle the question? Assume position axis is pointing up and the origin is on ground level.

Assume the cricket has a start speed of v0. It will go up, and the speed will diminish until it's zero, at which moment it starts falling down again.
The speed will diminish with g m/s per second. That is, v = vo - g.t

During a given time t, the distance traveled is x=x0+v0t-at²/2

These formula's should look quite familiar to you if you have to solve the question. There might be a difference in signs however, depending how you define the axis.

The way to go is to find out first how much time it takes to get to the top of the jump. At that moment, you know a lot of parameters for both equations, enough to conveniently plug one into the other.

From there, the rest should be simple.

See if you get further this way. If not, show us your way of reasoning, and we'll hint you further. Sorry for not just typing out the solution, but keep in mind we won't be there during tests either...

Good luck !

Leo
 
BronBurs said:
Heres the question: Some crickets can jump 21 cm, how long does it take the cricket to jump 7cm of that 21cm?
One way is to solve it backwards. Can you answer these questions:
(1) How long does it take something to fall 21 cm?
(2) How long does it take something to fall 14 cm?
 
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