Baseball acceleration physics

1. Sep 19, 2007

MIA6

1. A baseball is thrown vertically into the air with a speed of 24.0m/s. How long does it take to return to the ground?
I found out the height that it could reach was 29m, then I used the formula d=vit+at^2/2. ---> 29=24t+9.81t^2/2. But I got two answers. THey are both positive. Why? Is anything wrong?
2. A car traveling 90km/h decelerates at a constant 1.6m/s^2. Find out the distance it travels during the first and third seconds. Here it means the distance when the time interval is from the first to the third second, which is 2 seconds? OR it asks you the instantaneous velocity which is at exactly the first second, and exactly the third second.
btw, what do symbols like 'd0', 'a0' mean? they mean the distance and accelerations are 0? My teacher wrote these symbols. Maybe it's not formal.
Thanks.

Last edited: Sep 19, 2007
2. Sep 19, 2007

mgb_phys

1, The accelaration is negative, it acts in the opposite direction to the velocity ( otherwise it would keep going up for ever).

2, It is asking the distance travelled from t=0 to t=1 second and from t=2 to t=3 seconds.

3. Sep 19, 2007

MIA6

For the first one,how come I got two answers for time?
For the second one, the distance it travels the first second is 25m. For the third second, I got 25-2*1.6=21.8 Is that right?

Last edited: Sep 19, 2007
4. Sep 19, 2007

learningphysics

Why are you calculating the time to reach 29m? You need the time to return to the ground, ie 0m

5. Sep 20, 2007

MIA6

Can anyone give me a hint in how to do the second question?
btw, Is t up=t down in the first question?

Last edited: Sep 20, 2007
6. Sep 20, 2007

mgb_phys

Plot a graph of speed against time from 90km/h to 0.

Now work out the area under the graph between t=0 and t=1. That is the distance it moved. Now do the same for between t=2 to t=3

Ussually they would mean the displacement and accelaration at time or position 0.