While standing on a bridge 19.1 m above the ground, you drop a stone from rest. When the stone has fallen 3.40 m, you throw a second stone straight down. What minimal 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.

how long to fall 3.4 m ?

(1/2) (9.81) t^2 = 3.4
t = .833 seconds

how long to fall 19.1 meters ?
(1/2)(9.81) t^2 = 19.1
t = 1.97 seconds

so second rock fall for (1.97-.833) = 1.14 seconds

-19.1 = Vi t + (1/2)(-9.81) t^2
-19.1 = Vi (1.14) - (1/2)(9.81)(1.14)^2

1.14 Vi = -19.1 + 6.37

Vi = -11.2 m/s

To find the minimal initial velocity needed for the second stone to reach the ground at the same instant as the first stone, we can use kinematic equations.

Let's assume the time taken for both stones to reach the ground is t.

For the first stone:
Initial velocity (u1) = 0 (as it is dropped from rest)
Final velocity (v1) = ?
Acceleration (a1) = acceleration due to gravity = -9.8 m/s² (taking downward direction as negative)
Displacement (s1) = 19.1 m (since the stone is dropped from a height of 19.1 m)

We can use the formula: s1 = u1 * t + (1/2) * a1 * t²

Plugging in the values, we get:
19.1 = 0 * t + (1/2) * (-9.8) * t²
19.1 = -4.9t²

Now, let's consider the second stone:
Initial velocity (u2) = ?
Final velocity (v2) = ?
Acceleration (a2) = -9.8 m/s² (same as the first stone)
Displacement (s2) = 3.40 m (since it is dropped after a displacement of 3.40 m)

Using the same formula: s2 = u2 * t + (1/2) * a2 * t²

Plugging in the values, we get:
3.40 = u2 * t + (1/2) * (-9.8) * t²
3.40 = u2 * t - 4.9t²

Now, since both stones reach the ground at the same time, the time taken by both stones is the same (t).

By equating the expressions for the time, we can eliminate the variable t:
-4.9t² = 19.1
t² = 19.1 / -4.9
t² ≈ -3.898

However, since time cannot be negative, this result is not physically meaningful. It indicates that the second stone cannot be thrown from rest in order to reach the ground at the same instant as the first stone.

Therefore, it is not possible to find a minimal initial velocity for the second stone given the conditions described.