Why is the solution for time of flight multiplied by a factor of 2?

[VKnopp]

I have decided to self-teach myself A-Level Physics. I am confused about one particular step of a solution to a certain problem that the textbook provided.

Problem:

A stone is thrown vertically upwards at $15 ms^{-1}$. How long is the stone in the air until hitting the ground?

There are a few assumptions:

-air resistance is minimal
-displacement and velocity are positive upwards and negative downwards
-acceleration is always downward and therefore negative
-acceleration due to gravity is constant

Solution:

Time of flight is $2t$, twice time to maximum height. Maximum height is when final velocity $v=0$.

They use the classic equation.

$v=u+at$

$v=0$, $u=15 ms^{-1}$ and gravity is $9.8 ms^{-2}$.

$0=15+(-9.8)t$

Solving for $t$ yields $t=1.5306$. I understand all the way up to here.

They then multiply their solution for $t$ by a factor of $2$ to get the solution. I do not understand why they do this. Why do they do this step?

 

[Geofleur]

The time $t = 1.5306$ s is how long it takes the stone to reach its maximum height. The idea behind multiplying by 2 is that the whole process is symmetrical - it takes the same amount of time for the stone to fall back to the ground as it did for the stone to make the trip upward. You could divide the problem in two. The first problem would be to calculate the time for the trip up (and you have that solution already); the second problem would be to have the initial velocity be 0 and calculate the time for the velocity to reach -15 m/s. You will get $t = 1.5306$ s again.

 

[phinds]

Just as an aside, there is another VERY important assumption, and one that is normally made in such introductory problems, that you have not stated, but which actually should always be stated even though it often isn't. Can you think what it is?

 

[VKnopp]

The time $t = 1.5306$ s is how long it takes the stone to reach its maximum height. The idea behind multiplying by 2 is that the whole process is symmetrical - it takes the same amount of time for the stone to fall back to the ground as it did for the stone to make the trip upward. You could divide the problem in two. The first problem would be to calculate the time for the trip up (and you have that solution already); the second problem would be to have the initial velocity be 0 and calculate the time for the velocity to reach -15 m/s. You will get $t = 1.5306$ s again.

Of course. I get it know, thank you very much.

 

[VKnopp]

Just as an aside, there is another VERY important assumption, and one that is normally made in such introductory problems, that you have not stated, but which actually should always be stated even though it often isn't. Can you think what it is?

Is it gravity? I should assume that gravity pulls objects down at a rate of $9.8 ms^{-2}$.

 

[phinds]

Is it gravity? I should assume that gravity pulls objects down at a rate of $9.8 ms^{-2}$.

But you already have the 9.8 in your stuff, so no that's not it. Think about where the rock starts its travels.