Solve Kangaroo Jump Time Problem - Get Help Now

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In summary, the kangaroo will take twice as long to come down as it does to go up, so its total time in the air is 200 seconds.
  • #1
888
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hi,
i nead help on solving a problem can someone help me please?
if a kangaroo jumps a vertical height of 2.7m how long will it stay in the air before returning to the earth?

thanks
 
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  • #2
What have you tried? What formula do you think you would use?
 
  • #3
i think i would use xf=xi+vit+.5at^2
 
  • #4
i tried using it but i got so confused
there are other formulas i know but they don't have variable t for me to solve and there's vf=vi+at but i don't know vf
 
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  • #5
What will take longer? The time for the kangaroo to reach its maximum height, or the time for the kangaroo to drop from its maximum height back to Earth? What is the kangaroo's velocity when it is at its maximum height?
 
  • #6
it'll take longer to reach maximum height b/c of gravitational force and at max. height the vel is 0m/s
 
  • #7
so is the final vel. 0m/s? i thought the initial vel. was 0m/s?
 
  • #8
Yes, at max height the velocity is 0. The initial velocity can not be 0 or the kangaroo would not leave the ground.

It will not take longer to reach max height than it will to drop from max height back to the ground. They will be equal. This is very important because lots of the questions you will do depend on this.

If time up = time down, then you only need to solve for one of them and then times it by 2 to get your answer. You're looking for time. The formula you gave me is correct. Can you use algebra to re-write that formula so it looks like " t = ...something " ?
 
  • #9
but I am only finding the time that it last in the air so i don't need to multiply the answer by 2 rite?
 
  • #10
It depends on what you compute. I find it easiest to compute how long it will take something to drop from rest from a height of 2.7 meters. But you could compute how long it would take something to ascend from the ground to a height of 2.7 meters as well. The kangaroo's total time in the air will be the time it took to rise to max height + the time it took to drop back to the ground from max height. So if you figure out one of these, and the other one is equal, you just multiply by 2.
 
  • #11
but don't i need to know what vi is or else i have two unknown variables in the equation t and vi
 
  • #12
** edit. Ignore this post. It was meant for another thread. Sorry...

but if you write the equation for Earth, and set it equal to the equation for the Moon, things will cancel and you'll see this as a proportion rather than a formula that gives you an actual answer.

That's all your final answer was. It was a proportion. It didn't matter what vi was for Earth. The same vi on the Moon got you 6 times higher.
 
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  • #13
?? What does the moon have to do with this??

First, the initial velocity can't be 0 or the kangaroo will never get off the ground! The velocity of the kangaroo will be v= vi+ at at any time t (constant acceleration a so change in speed= at). The kangaroo will continue to go up until its velocity is 0 (of course, a= -9.8 m/s2 here). That is, the kangaroo will reach its greatest height when vi- 9.8t= 0 or t= vi/9.8. Yes, that depends on vi. Now put that for "t" in the height formula:xf= xi+ vit- 4.9t2. Taking xi= 0 (the kangaroo starts on the ground) and xf= 2.7 m, you have vi(vi/9.8)- 4.9(vi/9.8)2= 2.7. Solve that for vi and use vi to find the time t at which the kangaroo is at its highest point. As tony873004 told you, the kangaroo will take exactly as long to come down as to go up so its total time in the air is twice its time to the highest point.
 
  • #14
i'm having trouble rearranging the equation to get vi by itself and when i do i get the result of -518.62m/s as an answer for vi, can someone help me
 
  • #15
and when i plug it into the equation to get the time it turns out to be 100 and something which doesn't make sense
 
  • #16
HallsofIvy said:
?? What does the moon have to do with this??
.
oops. I'm getting my posts mixed up. This was supposed to be in reply to the 'gravity help' thread. sorry.
 
  • #17
it's okay tony873004 we all make mistakes =)
 

1. How do I solve the kangaroo jump time problem?

To solve the kangaroo jump time problem, you will need to use the formula (x2 - x1) / (v1 - v2), where x1 and x2 are the starting positions of the kangaroos and v1 and v2 are their respective speeds. This will give you the number of jumps it takes for the kangaroos to meet.

2. What are the variables in the kangaroo jump time formula?

The variables in the kangaroo jump time formula are x1 and x2, which represent the starting positions of the kangaroos, and v1 and v2, which represent their respective speeds.

3. Can I use the same formula for any numbers?

Yes, you can use the same formula for any numbers as long as they represent the starting positions and speeds of the kangaroos accurately. However, keep in mind that the formula will only give you the number of jumps it takes for the kangaroos to meet, not the actual time it takes.

4. How accurate is the kangaroo jump time formula?

The kangaroo jump time formula is accurate as long as the starting positions and speeds of the kangaroos are entered correctly. However, it does not take into account factors such as changes in speed or direction during the jumps, so it may not be completely accurate in real-life scenarios.

5. Are there any other methods to solve the kangaroo jump time problem?

Yes, there are other methods to solve the kangaroo jump time problem, such as using mathematical equations or creating a visual representation of the jumps. However, the formula (x2 - x1) / (v1 - v2) is the most commonly used and efficient method.

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