While standing on a bridge 15.0 m above the ground, you drop a stone from rest. When the stone has fallen 3.10 m, you throw a second stone straight down. What initial velocity must you give the second stone if they are both to reach the ground at the same instant? Take the downward direction to be the negative direction

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To solve this problem, we need to find the initial velocity required for the second stone so that it reaches the ground at the same time as the first stone.

First, let's find the time it takes for the first stone to reach the ground. We can use the equation of motion:

Δy = v0t + (1/2)gt^2

Where:
Δy = displacement (change in position)
v0 = initial velocity
t = time
g = acceleration due to gravity

In this case, the initial velocity (v0) of the first stone is 0 m/s since it is dropped from rest. The displacement (Δy) is the height of the bridge, which is -15.0 m since the downward direction is considered negative. The acceleration due to gravity (g) is approximately -9.8 m/s^2 (negative because it acts in the opposite direction to the positive y-axis).

Plugging these values into the equation, we have:

-15.0 m = 0 m/s * t + (1/2)(-9.8 m/s^2)t^2

Simplifying the equation, we get:

4.9t^2 - 15.0 = 0

This is a quadratic equation. Solving it will give us the time it takes for the first stone to reach the ground.

Using the quadratic formula:

t = (-b ± √(b^2 - 4ac)) / (2a),

where a = 4.9, b = 0, and c = -15.0, we find:

t = (√(0^2 - 4 * 4.9 * -15.0)) / (2 * 4.9) or t = - (√(0^2 - 4 * 4.9 * -15.0)) / (2 * 4.9)

Taking the positive value of t, we get:

t ≈ 1.22 seconds

Now, since the stones must reach the ground at the same time, we need to find the initial velocity (v0) of the second stone such that it falls the remaining distance of 15.0 m in the same time, 1.22 seconds.

We can use the kinematic equation again, with the final position of the second stone being -15.0 m (since it is also in the downward direction):

Δy = v0t + (1/2)gt^2

Plugging in the values:

-15.0 m = v0 * 1.22 s + (1/2)(-9.8 m/s^2)(1.22 s)^2

Simplifying:

-15.0 m = 1.22v0 - 7.1 m

Rearranging the equation:

1.22v0 = -7.9 m

Finally, solving for v0:

v0 ≈ -6.48 m/s

Therefore, the initial velocity that must be given to the second stone is approximately -6.48 m/s (negative to indicate the downward direction).