A woman on a bridge 84.5 m 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 6.00 m more to travel before passing under the bridge. The stone hits the water 4.00 m in front of the raft. Find the speed of the raft.

V = 0.48 m/s

Explanation:

In this case, we need to analyze the given data by parts.

At first, we know that the woman is on a height of 84.5 m of a river. She drops a stone thinking that she may hit the raft that is traveling with a constant speed. When the raft is 6 m near the bridge, the woman drops the stone, and the stone hits the water when the raft is still 4 m far of the bridge.

With this given data, we can calculate the distance covered by the raft, because is traveling at a constant speed:

X = 6 - 4 = 2 m

And as it's traveling at constant speed then:

X = V.t

We have the distance of the raft, but not the time it took to cover that distance. This time will be the same time that the stone took to hit the water, therefore, if we can determine the time of the rock, well be determining the time of the raft to cover the distance, and then, we can determine it speed.

To determine the time of the rock, as the stone is going on a free fall, with an innitial speed of 0, the flight time of the rock will be:

y = gt²/2 ---> solving for t

2y/g = t²

t = √2y/g

If g = 9.8 m/s, and replacing the data we have that the flight time of the rock is:

t = √2*84.5 / 9.8

t = 4.15 s

So the rock took 4.15 s to hit the water, and it's also the time that the raft took to cover the distance of 2 m, then, it's speed:

V = X/t

V = 2 / 4.15

Hope this helps

i believe the answer is a