For free fall problems what exactly does time (t) represent?

[physlexic]

Homework Statement

Whenever I solve a free-fall problem and it asks for the time, I never am sure what the time represents once I solve for it.
I understand when you throw a ball in the air and it comes back down, the time it takes to ascend up, is the same time it takes to descend down. Therefore, whenever I'm attempting a free-fall problem, and it asks for the time the ball was in the air, whatever I get for time, using a kinematics equation, I always multiply by 2 because as I stated before, you have to consider the same amount of time it ascends up + the same amount of time it descends down.
However, I will get problems wrong on my exam by multiplying by 2, and sometimes I won't multiply by 2 and I will get problems wrong because I didn't. Therefore, I just don't understand what time refers to in a free-fall problem.
Sorry for the long explanation but I had to include my confusion.
Here is an example of a problem where I multiplied by 2 and got it wrong:

"A ball is shot straight up from the surface of the Earth with an initial speed of 19.6 m/s. Neglect any effects due to air resistance, how much time elapses between the throwing of the ball and its return to the original launch point?
[/B]

Homework Equations

The Attempt at a Solution

V0 = 19.6 m/s
a = -9.80 m/s2
y = 0 m
t = ?I used equation
y = (v0)(t) + 1/2(a)(t^2)

solving for t I rearranged the equation to be

t = 2(y-v0)/a

t = 2[(0 m) - (19.6 m/2)] / (-9.8 m/s^2)
t = 4 s

Now what does this 4 seconds represent? Is it the total time the ball is in the air, or is it when the ball is ascending up and you have to remember that it descends the same amount of time as well, so multiply by 2? Because I assumed you had to multiply by 2 and left an answer of 8 seconds and got the problem wrong, it was 4 seconds.

However, another student used a difference equation (v = v0 + at) and got 2 seconds as his answer, and left it as 2, and got it wrong because he did not consider the ascending and descending time.

While typing this I think I just solved my own question...in order to realize what the time represents do I solve t for various other kinematic equations and see what other answers I get? The smaller time I get, then it's an indication that it's half the time it went up (or down) and therefore multiply by 2? [/B]

 

[ash64449]

Makes sense. Look at your method of approach.

Understand the equation.

you equated displacement to be 0. therefore the time t in your equation corresponds to the whole time. since the whole time,after the experiment is over,displacement is 0.

your kinematic equation gives relation between displacement and gives relation of time t taken for such a displacement to happen

 

[ash64449]

and you said your friend used v=v_0 + at

makes sense again.

he used only for the part of the journey. that is when the ball is thrown and when it is momentarily at rest upwards. that's why he equated v=0.

therefore that 't' corresponds to the time ball takes to reach that upward position.

solving you get t=2s. but this time is for half-way. so for full way multiply by 2.

therefore total time is 2*2=4s.

 

[ash64449]

there is still another way to solve this problem.

i take the convention that upward is positive and downward is negative.

for this you need to know an important result: when the ball reaches downward-starting point, speed is the same as the speed with which you throw the ball upwards.

but take care about the direction, direction is opposite.

according to my sigh convention, at the beginning of my experiment velocity is +u and at the end is -u.

well,when this happens,you complete the whole journey.. so when i put theses variables in velocity-time relation, the time 't' corresponds to the whole journey.

v=v_0 + at

-u=u-gt
-2u=-gt
therfore total time t=2u/g=(2*19.6)/9.8=4s.

 

[ash64449]

Your problem was at understanding the equations and the correspondence.

 

[ehild]

t is the elapsed time since the motion started. Imagine you have a stop watch and start it when the ball is thrown upward.
The velocity of the ball changes according to the equation
v=v₀-gt.
The speed decreases and becomes zero for an instant at $t=v_0/g$. The ball does not rise further, it is at the highest point of its trajectory. The velocity goes on decreasing further, so it becomes negative, and the ball descending.
The height changes with time as y=v₀t - g/2 t². When the ball returns to the ground y=0. At t=v_0/g the height is maximum. It is

$y_{max}=v_0 \frac {v_0}{g}-\frac{g}{2} \left(\frac {v_0}{g}\right)^2=\frac {v_0^2}{2g}$
Substitute twice of the rise time into the equation of y:
$y=v_0 \left(\frac {2v_0}{g}\right)-\frac{g}{2}\left(\frac {2v_0}{g}\right)^2 =0$,
the ball reached the ground (y=0). It ascended for t=v_0/g and descended for the same time to reach the ground.

You get the total time in air also by solving the quadratic equation $v_0t-\frac{g}{2}t^2=0$ What roots you get?

 

[tommyxu3]

I used equation
y = (v0)(t) + 1/2(a)(t^2)

solving for t I rearranged the equation to be

t = 2(y-v0)/a

How do you rearrange the equation to get t? Moreover, the $y$ you write should be the displacement and I'm not sure what you presented in yours.
I think the example you posted not a free fall, for a free fall should start from static. The problem I think is just a motion under the constant gravitation hence also a motion with constant acceleration.
So the time the example wants you to calculate has been clearly articulated in it:

between the throwing of the ball and its return to the original launch point?

And I suggest you deal with it more easily. $a=-9.8m/s^2, v_0=19.6m/s,$so it takes $2~ sec$ to reach the highest, and then therefore all the process takes $4~ sec.$

 

[tommyxu3]

I'm sorry for neglecting ehild's last question... I just had the same too.