A stone is dropped from rest from a bridge that is 45 m above a river. A second stone is thrown down 1.o s after the first stone is dropped. The two stones strike the water simultaneously. What was the initial velocity of the second stone?

The first stone:

45 = 4.9t^2, t = 3.03 s. = Fall time.

The 2nd stone:
h = Vo*t + 0.5g*t^2.
45 = Vo*(3.03-1) + 4.9(3.03-1)^2,
45 = Vo*2.03 + 4.9*2.03^2. Vo = ?.

To solve this problem, we need to consider the motion of both stones separately.

Let's start with the first stone that was dropped from rest. The distance it falls from the bridge to the water is 45 m. We can use the kinematic equation:

d = (1/2) * g * t^2

where:
d is the distance it falls (45 m),
g is the acceleration due to gravity (9.8 m/s^2),
t is the time it takes to fall.

Rearranging the equation, we can find the time it takes for the first stone to fall:

t = sqrt(2d/g) = sqrt(2 * 45 / 9.8) ≈ 3 s.

Now, let's consider the second stone. It was thrown down 1.0 s after the first stone was dropped. This means the second stone has been falling for 2 seconds when the first stone hits the water (3 s - 1 s).

Since both stones hit the water simultaneously, the total time for the second stone to fall from its initial position to the water is 2 s.

Using the same kinematic equation, we can find the initial velocity of the second stone:

d = (1/2) * g * t^2

Rearranging the equation, we can solve for the initial velocity (v0):

v0 = (d * 2 / t^2)^(1/2)

Substituting the known values:
v0 = (45 * 2 / 2^2)^(1/2) = (45 * 2 / 4)^(1/2) = (90 / 4)^(1/2) = √22.5 ≈ 4.74 m/s.

Therefore, the initial velocity of the second stone is approximately 4.74 m/s.

To determine the initial velocity of the second stone, we can use the kinematic equations of motion. We will consider the motion of the second stone since its initial velocity is unknown.

Let's define some variables:

h = height of the bridge (45 m)
t1 = time taken for the first stone to reach the water (unknown)
t2 = time taken for the second stone to reach the water (1.0 s)

Using the kinematic equation
h = ut + (1/2)gt^2

For the first stone, which is dropped from rest, the initial velocity (u) is 0 m/s. Therefore, the equation becomes:
h = (1/2)gt1^2

For the second stone, we need to find its initial velocity (u2). The equation becomes:
h = u2t2 + (1/2)gt2^2

Now we have two equations:

(1) h = (1/2)gt1^2 --- Equation 1
(2) h = u2t2 + (1/2)gt2^2 --- Equation 2

Since both stones strike the water simultaneously, we can equate t1 and t2. So we have:
t1 = t2

Substituting this in Equation 2, we get:
h = u2t1 + (1/2)gt1^2

Now we have two equations with two unknowns:
h = (1/2)gt1^2 --- Equation 1
h = u2t1 + (1/2)gt1^2 --- Equation 2

Subtracting Equation 1 from Equation 2:
u2t1 = 0

From the above equation, we can determine that either u2 or t1 is equal to 0. Since time cannot be 0 in this scenario, it means that the initial velocity of the second stone (u2) is 0 m/s.

Therefore, the initial velocity of the second stone is 0 m/s.