Basic physics energy problem?

h1 = 0.03m

h2 = 0.887m

d1 = 1m

d2 = ? –– where the ball lands

t = 1.69s –– the time it takes to travel across the table (1m)

Using conservation of energy

Ui + Ki = Uf + Kf

mgh1 + 0 = 0 + ½mv²

mgh1 = ½mv²

√2gh1 = v

0.7668 = v –– velocity at the bottom of the ramp

This velocity will be less when it reaches the edge of the table due to friction. First find it's acceleration, then use that to find it's final velocity at the edge.

d1 = Vi*t + ½at²

2(d1 - Vi*t) / t² = a

2[1 - (0.7668)(1.69)] / 1.69² = a

-0.2072 = a

Velocity at the edge of the table is

Vf = Vi + at

Vf = 0.7668 + (-0.2072)(1.69)

Vf = 0.4166

The time it takes for the ball to fall is

h2 = Vi*t + ½at²

h2 = 0 + ½at²

√(2h2 / a) = t

0.4255 = t

The location where it lands is

d = vt

d2 = (0.4166)(0.4255) = 0.1773m

d2 = 17.72cm

this is what it should be if t = 1.69.

Sometimes, what you observed and what you worked out on paper don't agree perfectly because of errors in measurements. If you observed d2 to be 23cm, then the time you should've measured is t = 1.53 based on the math. Which is only a difference error of 0.16 seconds. So you can see, even a small error will still give you different, but close results.

What I think:

First off, like you said, 0.5322 is wrong. I'm guessing you also forgot to take into account the ball slows down a little before it drops. So you did...

d2 = (0.5322)(0.4255) = 0.226m = 23cm

which is wrong, but you got 23cm anyway when you tested it, which is purely coincidental.