Question

I have a question from my book that i have no idea how they came up with the answer. If anyone could explain i would be grateful.
"A brick is thrown vertically upward, reaching the top at 6m. You are standing at 5m high and catch it. How long does it take for the brick to reach this you?"

Answer 1

I'd say calculate the time it takes for an object to fall 6 meters then add to that the amount of time it takes for an object to fall 1 meter.

Answer 2

ok, i'll try that....

Answer 3

What is the answer that they gave?

Answer 4

they gave 0.65s

Answer 5

Okay the formula is d = 1/2 * g * t^2 where g = 9.80665 m/s2
Solving the formula for t, we get
t^2 = d / (1/2 *g) or
t = sqroot (d / 4.903325)
For the 6 meters
t = sq root (6/4.903325)
1.106191419 seconds

To fall 1 meter
t = sq root (1/4.903325)

=0.4516007558 seconds

Adding both times

1.5577921748 seconds
or about 1.56 seconds
and that is not what they said is it?

Answer 6

RIght? I never came even all that close to the answer of .65s!!!! I got EXACTLY what you got!

Answer 7

ok, so maybe not *exactly* (rounding and all), but you can see my frustration

Answer 8

Well, I don't know what to say. The formula is the same and the value of g is the same.

Answer 9

Im thinking the book might very well be wrong. Hmmmmm, thanks for confirming what i got. I really appreciate your efforts!!!!

Answer 10

Okay I'm glad to help out. :-)