Question

A woman on a bridge 95.6 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 4.84 m more to travel before passing under the bridge. The stone hits the water 1.56 m in front of the raft. Find the speed of the raft.

Answer 1

Answer:

The speed of the raft is 0.74 m/s

Explanation:

In this exercise we have two moments, the first, before the stone is released and the second, when the stone hits the water. the distance traveled by the raft under its free fall is calculated with the following equation:

x = 4.84 - 1.56 = 3.28 m

The distance is equal to:

x = v*t (eq. 1)

the distance in y direction is equal to:

y = u*t + (a*t^2)/2

if the stone initially at rest, u = 0

y = (a*t^2)/2

Clearing t:

t = (2*y/a)^1/2 (eq. 2)

if we substitute equation 2 in equation 1:

x = v*(2*y/a)^1/2

Clearing v:

v = x*(a/2*y)^1/2 = 3.28*((-9.8)/(2*(-95.6))))^1/2 = 0.74 m/s

Answer 2

Answer:

The speed of the raft is 1.05 m/s

Explanation:

The equation for the position of the stone is as follows:

y = y0 + v0 · t + 1/2 · g · t²

Where:

y = height of the stone at time t

y0 = initial height

v0 = initial speed

t = time

g = acceleration due to gravity

The equation for the position of the raft is as follows:

x = x0 + v · t

Where:

x = position of the raft at time t

x0 = initial position

v = velocity

t = time

To find the speed of the raft, we have to know how much time the raft traveled until the stone reached the river. For that, we can calculate the time of free fall of the stone:

y = y0 + v0 · t + 1/2 · g · t²      (v0=0 because the stone is dropped from rest)

If we place the origin of the frame of reference at the river below the bridge:

0 m = 95.6 m - 9.8 m/s² · t²

-95.6 m / -9,8 m/s² = t²

t = 3.12 s

We know that the raft traveled (4.84 m - 1.56 m) 3.28 m in that time, then the velocity of the raft will be:

x/t = v

3.28 m / 3.12 s = v

v = 1.05 m/s