While standing on a bridge 15.0 m above the ground, you drop a stone from rest. When the stone has fallen 5.00 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.

First calculate the time to fall Y = 5.00 m and Y = 150 m for the first (dropped) stone. t = sqrt(2Y/g)

The times are 1.02 s and 5.53 s

Then calculate the speed V the second stone must be thrown to reach the ground at the same time.

150 = V*(t-1.02) + (g/2)(t-1.02)^2

Solve for V

To find the initial velocity of the second stone, we can use the kinematic equation:

\(h = v_0t + \frac{1}{2}gt^2\)

Where:
- \(h\) is the height (distance) fallen
- \(v_0\) is the initial velocity
- \(t\) is the time taken
- \(g\) is the acceleration due to gravity, which is approximately -9.8 m/s² in the downward direction

For the first stone, which is dropped from rest, the equation becomes:

\(5 = \frac{1}{2}(-9.8)t^2\)

Simplifying, we can solve for \(t\):

\(5 = -4.9t^2\)
\(t^2 = \frac{5}{-4.9}\)
\(t = \sqrt{\frac{5}{-4.9}}\)
\(t \approx 0.451\) seconds

Now, we can use this time to find the initial velocity of the second stone. Since both stones are supposed to reach the ground at the same time, the time for the second stone will also be 0.451 seconds.

Using the same equation for the second stone, with the additional initial velocity term, we have:

\(15 = v_0(0.451) - 4.9(0.451)^2\)

Now, we can rearrange the equation and solve for \(v_0\):

\(v_0 = \frac{15 + 4.9(0.451)^2}{0.451}\)

Evaluating the equation, we find that the initial velocity of the second stone must be:

\(v_0 \approx \frac{15 + 4.9(0.451)^2}{0.451} \approx 16.3\) m/s (rounded to one decimal place)

Therefore, to throw the second stone such that it reaches the ground at the same time as the first stone, you must give it an initial velocity of approximately 16.3 m/s in the downward direction.