Posted by **mary** on Sunday, December 28, 2008 at 12:12am.

A woman on a bridge 90.0m high sees a raft floating at a constant sped on the river below. She drops a stone from rest in an attempt to hit the raft. The stone is released when the raft has 6.00m more to travel before passing under the bridge. The stone hits the water 2.00m in front of the raft. Find the speed of the raft.

I made a table with Y:-90m, A:9.8m/s^2, Vo: 0

So since y+ Vot+1/2at^2

90= 0+ ½(9.8) t^2

T+ 4.29

I figured the raft moves 4.00m.

So

4.00m/4.49s =.89 m/s

But the answer in the back of the book is .932m/s

What did I do wrong?

- Physics -
**drwls**, Sunday, December 28, 2008 at 1:49am
What you did wrong was forget that you had (correctly) calculated the time to reach the water as 4.29 s. You then changed it to 4.49 s.

The raft moves 4 m in the time it takes the stone to fall 90 m. That time T is given by the equation

(g/2)T^2 = 90

T = sqrt(180/g) = 4.29 s

V = 4/4.29 = 0.932 m/s

- Physics -
**~christina~**, Sunday, December 28, 2008 at 1:50am
The problem is that you divided the distance traveled by the raft by ..4.49s.

I don't know where you got 4.49, but using the original time obtained by you with the time it takes for the rock to hit the water, would give you the answer in the book.

(time it takes for stone to hit water is same time traveled by raft)

- Physics -
**mary**, Friday, January 2, 2009 at 10:09pm
thanks!

- Physics -
**Precious**, Wednesday, January 18, 2012 at 5:46am
A woman on a bridge 75.0m high sees a raft floating at a constant speed on the river below. She drops a stone from rest in an attempt to hit the raft. The stone is released when the raft has 7.00m more to travel before passing under the bridge. The stone hits the water 4.00m in front of the raft. Find the speed of the raft.

- Physics -
**Jec**, Tuesday, September 16, 2014 at 8:34am
How did you get the time,?

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