Solve Kangaroo Jump Time Problem - Get Help Now

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Homework Help Overview

The discussion revolves around a physics problem involving a kangaroo jumping to a height of 2.7 meters and the time it remains in the air. The subject area includes kinematics and projectile motion.

Discussion Character

  • Exploratory, Conceptual clarification, Mathematical reasoning, Problem interpretation

Approaches and Questions Raised

  • Participants discuss various formulas related to motion, including the use of kinematic equations. Questions arise about the initial and final velocities, the time taken to reach maximum height versus the time taken to fall back down, and how to rearrange equations to isolate variables.

Discussion Status

Participants are actively engaging with the problem, exploring different interpretations and approaches. Some guidance has been provided regarding the relationship between ascent and descent times, and the importance of understanding initial velocity. However, there is no explicit consensus on the correct approach or solution yet.

Contextual Notes

There are indications of confusion regarding the variables involved, particularly the initial velocity and how it relates to the time of flight. Some participants express difficulty in rearranging equations and obtaining meaningful results.

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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What have you tried? What formula do you think you would use?
 
i think i would use xf=xi+vit+.5at^2
 
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
 
Last edited:
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?
 
it'll take longer to reach maximum height b/c of gravitational force and at max. height the vel is 0m/s
 
so is the final vel. 0m/s? i thought the initial vel. was 0m/s?
 
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 " ?
 
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.
 
Last edited:
  • #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 =)
 

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