While standing on a bridge 30.0 m above the ground, you drop a stone from rest. When the stone has fallen 3.20 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.

d1 = 0.5g*t^2 = 30-3.2 = 26.8.

4.9t^2 = 26.8
t^2 = 5.47
t = 2.34 s. = Fall time of 1st stone.

The 2nd stone must fall 30 meters in
2.34 s.

d2 = Vo*t + 0.5g*t^2 = 30 m.
Vo*2.34 + 4.9(2.34)^2 = 30
2.34Vo + 26.83 = 30
2.34Vo = 30-26.83 = 3.17
Vo = 1.35 m/s. = Initial velocity of 2nd stone.

To solve this problem, we can use the equations of motion to calculate the time it takes for each stone to reach the ground. Since the first stone is dropped from rest, its initial velocity is zero. The second stone, however, is thrown downward, so we need to find its initial velocity.

Let's break down the problem step by step:

1. Calculate the time it takes for the first stone to reach the ground:
We can use the equation for free-fall motion: h = (1/2)gt^2, where h is the distance fallen, g is the acceleration due to gravity (approximately 9.8 m/s²), and t is the time. Rearranging the equation, we have t = sqrt(2h/g).

Substituting the given values, h = 30.0 m, we find:
t₁ = sqrt(2 * 30.0 m / 9.8 m/s²) ≈ 2.19 s

2. Find the velocity of the first stone just before it hits the ground:
The final velocity of an object in free fall can be calculated using the equation: v = gt. Since the initial velocity is zero, the final velocity is just the velocity just before hitting the ground. So, v₁ = g * t₁.

Substituting the known values, we have:
v₁ = 9.8 m/s² * 2.19 s ≈ 21.46 m/s

3. Determine the initial velocity of the second stone:
To ensure that both stones hit the ground at the same time, the second stone needs to have the same time of flight as the first stone. Therefore, the initial velocity of the second stone should be such that it will reach the ground after the same time interval as the first stone.

Since the initial position of the second stone is 3.20 m below the initial position of the first stone, we need to calculate the time it takes for this stone to fall this distance and subtract it from the total time of flight.

Using the same equation as before, t = sqrt(2h/g), we can find the time it takes for the second stone to fall 3.20 m:
t₂ = sqrt(2 * 3.20 m / 9.8 m/s²) ≈ 0.80 s

Subtracting this time from t₁ gives us the time of flight for the second stone:
t_total = t₁ - t₂ ≈ 2.19 s - 0.80 s ≈ 1.39 s

Now, we can find the initial velocity (v₂) of the second stone using the equation: h = v₀t + (1/2)gt².
Substituting the known values, h = -30.0 m (negative because the stone is thrown downward) and t = t_total, we can solve for v₀:
-30.0 m = v₀ * 1.39 s + (1/2) * 9.8 m/s² * (1.39 s)²

Rearranging the equation and solving for v₀, we have:
v₀ ≈ -30.0 m - 0.5 * 9.8 m/s² * (1.39 s)² / 1.39 s
v₀ ≈ -30.0 m + 9.8 m/s² * 0.9699 s
v₀ ≈ -30.0 m + 9.5 m/s
v₀ ≈ -20.5 m/s

Therefore, in order for both stones to reach the ground at the same instant, the second stone must be thrown with an initial velocity of approximately -20.5 m/s.