# Why does an object fall in the same time as it takes to reach a peak

mikeza

## Homework Statement

A kangaroo jumps a height of 2.8m. How long was it in the air?

## Homework Equations

v=v0+at

x=x0+v0t+(1/2)at^2

v^2=v0^2+2aΔx

## The Attempt at a Solution

-2.8 = -4.9t^2
.57 = t^2
t = .76 s (down)

0 = v0^2+2(9.8)(2.8)
54.88 = v0^2
v0 = 7.408

0 = 7.408-9.8t
t = .76 (up)

t = 1.5 sec

SO...
I know I have the answer right I just don't understand how that works conceptually - that if something has a initial velocity upwards it takes the same time as if the initial velocity was 0 downwards. How can something traveling the same distance (just one negative and one positive) with the same acceleration but different initial velocities take the same amount of time out to the 9th decimal??

QuarkCharmer
The thing initially has some kinetic energy. As it moves up, that kinetic energy is converted into gravitational potential energy, and then as it comes back down it is converted back into kinetic energy. Since energy is conserved, it would make sense that the initial velocity before leaving the ground is the same as just before it touches back down.

It doesn't work that way in the real world though, what with air resistance doing negative work on the way up and the way down, among other things.

mikeza
Thanks a bunch. Can't wait to teach my class tomorrow... my teacher doesn't tell us anything.

Homework Helper
Gold Member
Excellent observation, though you are not the first to make it. In fact it is much discussed you will later find.

The equations you quote are applications of Newton's equations of motion. These are reversible, i.e. if you exactly reverse the direction of every particle is travelling giving it the same velocity but in the exact opposite direction it will retrace the exact path it came by. People talk of substituting -t for t in equations and the solution is sort of the same in both cases but with time reversed.

To a extent this happens and can be observed in reality - as you have just done really. The reason it does not work in practice is also much discussed, and is that if the situation is more than so much complicated, you cannot realise that reversal. That is if you drop a drop of ink into water, according to the equations (which we do not have to solve, it is just a general property they have) if you could get behind each ink particle and for that matter all the water molecules, and push them back so as each have exactly equal opposite velocity they were travelling in that instant, then they would retrace their motions in reverse, reconcentrate together into an ink drop which would come out of the water back into the dropper. But you can see the 'if you could' is not a reasonable expectation, which explains why you will never see this happen.

If you are starting science expect to hear more of this.

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Homework Helper

## Homework Statement

A kangaroo jumps a height of 2.8m. How long was it in the air?

## Homework Equations

v=v0+at

x=x0+v0t+(1/2)at^2

v^2=v0^2+2aΔx

## The Attempt at a Solution

-2.8 = -4.9t^2
.57 = t^2
t = .76 s (down)

0 = v0^2+2(9.8)(2.8)
54.88 = v0^2
v0 = 7.408

0 = 7.408-9.8t
t = .76 (up)

t = 1.5 sec

SO...
I know I have the answer right I just don't understand how that works conceptually - that if something has a initial velocity upwards it takes the same time as if the initial velocity was 0 downwards. How can something traveling the same distance (just one negative and one positive) with the same acceleration but different initial velocities take the same amount of time out to the 9th decimal??

It may have been deliberate, but the "give away" equation you did not list [or was not stated] is

x = (vo + vf)/2 x time.

On the way up, the object starts at some velocity and finishes with zero velocity, thus covering a certain distance on the way. The time was how ever long it takes to change your velocity from the original value to zero, while acceleration is g.
On the way down, it has to cover the same distance, under the influence of the same acceleration [g] so will start at velocity 0, and finally reach the same speed as it was projected at, but travelling down not up.