How Long Was the Kangaroo in the Air?

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Homework Help Overview

The discussion revolves around a physics problem involving kinematics, specifically the motion of a kangaroo jumping to a height of 2.8 meters and determining the total time it was in the air. Participants are exploring the relationships between initial velocity, final velocity, acceleration, distance, and time.

Discussion Character

  • Exploratory, Conceptual clarification, Mathematical reasoning

Approaches and Questions Raised

  • Participants discuss the need to break down the problem into components and consider initial and final velocities, as well as acceleration due to gravity. There are attempts to apply kinematic equations to find the time of flight, with some questioning the assumptions about the initial conditions of the jump.

Discussion Status

Some participants have provided guidance on using kinematic equations to calculate the time to reach the peak of the jump and the overall time in the air. There is recognition of the need to multiply the time to the peak by two to account for the descent. Multiple interpretations of the problem setup are being explored, but no explicit consensus has been reached.

Contextual Notes

Participants note that the problem provides limited information, specifically regarding the initial velocity of the kangaroo and whether it starts from rest. This has led to varied approaches in solving the problem.

Amber
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Why can't I understand simple vector questions? I've just started a higher physics and that's what we're learning.

I'm stuck with this question:

A kangaroo was seen to jump to a vertical height of 2.8m. How long was it in the air?

I have the answer at the back of the book, but I don't know how to arrive at the answer! Please help!

Do I need to split it up into the components:

u - initial velocity
v - final velocity
a - acceleration
s - distance
t - time

I know the equations, but when I plug the numbers in, it doesn't come out as the right answer!

:confused:
 
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is that all it says, must say how much Vi it has, or does it start from rest.
If it starts from rest you find the time it takes it to get to the top and then multiply by 2 and that's the total time, using the equation x=Vit + (1/2)at^2.
 
Last edited:
two equations to be used here,
1>
Lets use the third kinematic eqn...
(using g down as positive sign so when direction of motion is upwards g is negative)
that gives us,
v^2=u^2-2gs

when it reaches maximum point , the final velocity becomes 0,
this gives v=0 and that gives us,
u^2 = 2gs
** we can calculate u since we know g and s

2>
now let's use the second kinematic equation,
s=ut+1/2at^2
from 1 we got 'u'
and we know a=-g and s=2.8
rearrange the equation and u get a quadratic in t
solve it to find t

But this t is the time to taken to reach the top we need the overall time which is just 2*t.

-- AI
 
I guess your right, looks like I messed up :frown: . Gj though. One more thing how does the kangaroo manage to jump up and straight down?, he must land a little to the side or sometin.
 
Why is that?
 
nm, sometimes i don't even know what I am typing but I just type it. :-p
 
That's all the information I'm given.

The answer is 1.5s, and I still don't know how to reach it! :frown:
 
Okay we have that
Vf=0 m/s
a=-9.81 m/s^2
x=2.8m

So in the equation

Vf^2 - Vi^2 = 2ax
0 - Vi^2 = 2 (-9.81)(2.8)
-Vi^2 = -54.936
Vi = 7.41m/s

Now in the equation

a = (Vi - Vf) / 2
-9.81 = (0-7.41) / t
-9.81t = -7.41
t = .76s

so that's only to the top so we multiply by 2, .76 * 2 = 1.52 s, which is the total time.
 
cdhotfire said:
so that's only to the top so we multiply by 2, .76 * 2 = 1.52 s, which is the total time.

That's what I didn't do! I now understand why I have to multiply it by two, I didn't think about that earlier! Thank you so much. :smile:
 
  • #10
No problem, glad I could help yout out. :cool:
 

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