# Two Dimensional Kinematics

## Homework Statement

In the photograph , suppose the cart that launches the ball is 13 cm high. The vertical distance from the lowest ball to the highest is about two cart heights or 26 cm.

Estimate the time interval between successive stroboscopic exposures.

Y=y0+V0t+1/2at

## The Attempt at a Solution

.26=0+2.3t+1/2*-9.81t^2
.26=2.3t-4.905t^2
-4.905t^2+2.3t-.26=0

t=-2.3+-sqrt 2.3^2-4*-4.905*-.26/-9.81
t=-2.3+-sqrt 5.29+19.62*-.26/-9.81
t=-2.3+-sqrt 5.29-5.1/-9.81
t=-2.3 +- sqrt .19/-9.81
t= 2.3+- .44/-9.81
t= .19, .28
t= .19/3= .06

Delphi51
Homework Helper
I'm not sure I understand what is happening in that diagram. I don't understand your initial velocity of 2.3.

Would it be correct to say the ball falls 26 cm from the turnover point in 3 flashes? If so you can find the answer quickly by using y = .5*g*t^2 with y and g both positive.

Well, I know the initial velocity is right, its a two part problem and that was the answer to part A, as for the diagram it's multiple pictures of a single ball being launched in an arc shape. I need to find the time between each picture. I tried finding the time to reach the max height, and dividing that by 3 (since there's three gaps between the first picture and that picture)

Delphi51
Homework Helper
Okay, your work all looks good based on the 2.3 initial velocity. I got slightly different answers using
t = sqrt(2*.26/9.81) = .23 s for three flashes during the fall. I think the difference is due to your 2.3 being slightly inaccurate. If you use v^2 = 2ad = 2g*.26 you get v = 2.26 instead of 2.3.

The rule I'm used to for accuracy is to keep an extra digit in all calculations, rounding only the final answer when you report it.

mmkay, i tried that, and I also got .23. But .23/3 gave me .063... and the answer was still incorrect, am I dividing by the right number of time intervals?

Thanks for your help Delphi, but I figured it out myself, the computer just wanted it in scientific notation for some reason >_>