Displacement Problem 2

1. Aug 25, 2009

cyspope

1. The problem statement, all variables and given/known data

A physics student trhow a ball vertically into the air. A fellow student determines with a stopwatch that the ball is in the air for a total of 3.56s. (a) with what velocity was the ball thrown? (b) How high did it rise?

3. The attempt at a solution

Vo = ?
Vf = 0
t = 3.56
yo = ?
yf = ?
a = ?

I don't know what equation I should use and how to set up the equations.

2. Aug 25, 2009

gabbagabbahey

To start with, after the ball is thrown, what force(s) are acting on it? So $a=$____?

3. Aug 25, 2009

cyspope

the gravity? so a = 9.8 m/s² right?

4. Aug 25, 2009

gabbagabbahey

I'd use a negative sign, since gravity will accelerate the ball downwards, but yes.

So now you have $v_f=0$, $\Delta t=t_f-t_i=3.56 \text{s}$ and $a=-9.8 \text{m/s}$ and you want to find $v_0$....Can you think of an equation that relates the quantities $v_0$, $v_f$, $a$ and $\Delta t$?

5. Aug 25, 2009

cyspope

a= (vf-vi)/(t) is this it

so,
-9.8m/s² = (0-vi) /3.56s
so vi = 34.888?
is this the answer for (a)?

Last edited: Aug 25, 2009
6. Aug 25, 2009

gabbagabbahey

Well, that equation applies to constant acceleration problems, and in this problem the acceleration is constant....so, yes

Now rearrange that equation to solve for 'vi' and then plug in the numbers and bingo....

7. Aug 25, 2009

cyspope

a= -9.8 m/s²
vi= 34.888
vf= 0
t = 3.56
this is what I've found so far based on what you said,

In order to find (b), should I use Uniform motion with constant acceleration, which is d=vi*t+(1/2)*a*t² or do I have to use some other equation?

8. Aug 25, 2009

gabbagabbahey

Hold on, I just re-read the question and realized that the 't' you are given is the total time that the ball is in the air, not the time it takes to reach the top of its motion (where vf would equal zero), so your calculation for (a) is incorrect.

Instead, break the motion into two sections. say the ball is thrown at time $t_0$, rises to a maximum height $d$ at time $t_1$ and then falls back to earth at time $t_2$.... you know that

$a=-9.8\,m/s^2$
$t_2-t_0=3.56\, s$
$v_1=0$ (since the speed at the very top of the motion is zero)

And you want to find (a)$v_0$ and (b)$d$

Can you think of which equations you might use here?

9. Aug 25, 2009

cyspope

I think I still have to use "average velocity and acceleration" equation which is, a=(vf-vi)/t.
Since you told me the calculation for (a) is wrong, can you please tell me which equation I need to use.

For (b) I think I need to use "Uniform motion with constant acceleration" formula , which is d=vi*t+(1/2)*a*t²